PrepShorts · Study sheet · Class 9 Mathematics · Chapter 4, Exploring Algebraic Identities
Chapter 4 · Exploring Algebraic Identities
(a + b + c)² by substitution, and the square that proves it
This video could not be loaded. Reload the page to try again.
Sign in with Google10 min.
Keep your place in this chapter — sign in, it’s free.Sign in
A new identity does not need a new proof if you can turn it into an old one in disguise. Do not start again — change a name.
The idea
A new identity does not need a new proof if you can make it an old one in disguise. Write d for b + c and the three-letter square collapses into the two-letter square you already proved; unwind d and the six terms fall out, with nothing new assumed. The drawn square then does something the algebra cannot: it shows why the answer has the shape it has — three squares along the diagonal, one for each letter, and every pair of letters twice, because each pair meets in two cells. And the section's second half turns the same idea around: an identity earns its keep when it converts a hard computation into an easy one, which is what Śhrīdharāchārya's rearrangement of a² − b² was for.
What you should be able to do
- Obtain (a + b + c)² by replacing b + c with a single letter and applying the two-term identity
- Unwind the substitution and collect the six terms of the expansion
- Write the result in the order that makes it memorable — the three squares first, then the three doubled products
- Label the nine pieces of a supplied 3 × 3 partitioned square so that the drawing represents the identity
- Explain from the figure why each pair of letters contributes twice while each single letter contributes once
- Square a three-digit number by splitting it into hundreds, tens and units
- Choose, for a given number, which of the three identities makes the squaring easiest, and say why
- Rewrite a² − b² = (a + b)(a − b) as a rule for computing a², and use it on a number ending in 5
- Justify that rewriting by cutting a square, turning one strip and reassembling
- Decide whether a proposed three-square statement is an identity, and correct it if it is not
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| identity | an equality that holds for every value of every letter in it | printed in this chapter (§4.2, p. 70) |
| geometrical model | a drawing whose areas stand for the terms of an algebraic statement | printed in this chapter (Fig. 4.4 caption, §4.4, p. 76) |
| substituting | putting one expression in place of a letter throughout | printed in this chapter (§4.7, p. 85; the move itself is made on p. 75, where the page says it is replacing b + c) |
| area | the surface a region covers, in square units | printed in this chapter (§4.2, p. 69) |
| square of side | the square drawn on a stated length | printed in this chapter (§4.2, Fig. 4.2 caption, p. 69) |
| Śhrīdharāchārya | Indian mathematician credited on p. 77 with the squaring method built on a² − b² | printed in this chapter (§4.4, p. 77) |
| factor | an expression that divides another exactly | printed in this chapter (§4.3, p. 72) |
| coefficient | the numerical multiplier of a term | printed in this chapter (§4.6, p. 81) |
| cross terms | the doubled products 2ab, 2bc, 2ca — the pieces that are not squares | an added term; the chapter lists these terms and gives them no collective name |
| diagonal cells | the three pieces of Fig. 4.4 that are squares rather than rectangles | an added phrasing, not printed in this chapter |
| dissection | a proof that cuts a figure up and reassembles the pieces | an added term, not printed in this chapter |
Where people slip up
- "A new identity needs a new proof." It needs a new proof only if you cannot reduce it to an old one. The substitution d = b + c is the cheapest proof in the chapter and it assumes nothing beyond the two-letter identity.
- "(a + b + c)² = a² + b² + c²." The same error as the two-letter case, three times over. The figure makes the missing pieces visible: six of the nine cells are not squares at all.
- "There should be three cross terms, one per letter." There are three, but one per pair, not one per letter — ab, bc, ca. With four letters there would be six. Count the off-diagonal cells and the rule becomes obvious.
- "2ca and 2ac are different terms." They are the same term; the page writes the second form because it reads round the triangle a → b → c → a. Say so once, or students will look for a seventh term.
- "With a minus in the middle, (a − b + c)² needs its own identity." It does not. Take b as negative and the same six-term expansion applies, which is what makes Q2 (v) and Q3 (ii) tractable.
- "Q4 must be an identity, because it is printed in the book." It is printed as a question. One substitution — a = b = c = 1 — refutes it. This is where the chapter's earlier lesson about a single counter-case pays off.
- "The three-term identity is always the best tool for a three-digit square." For 117 the two-term minus identity is faster. Exercise Set 4.3 Q1 is explicitly a question about which tool to reach for.
- "a² = (a + b)(a − b) + b² is a different fact from a² − b² = (a + b)(a − b)." It is the same fact with one term moved. What changes is what it is for: the first form computes squares, the second factorises differences.
- "Fig. 4.5 shows areas being lost." Nothing is lost. One strip is turned through a right angle and re-laid; the b-by-b corner is the only piece that stays where it was, and it is the whole reason for the +b².
Ask your teacher a person
Your teacher reads this and writes back, usually within a day. For an instant answer, use Ask the video in the sidebar.
Your class sees the question and the answer. Only your teacher sees that it was you.
No questions on this topic yet.
Worked answers to this chapter’s exercises · this video explains Exercise Set 4.3 Q1, Exercise Set 4.3 Q3, Exercise Set 4.3 Q4
Transcript1,431 words
You know what to do with the square of a sum of two things. Now somebody hands you three. a plus b plus c, all squared. The honest first thought is that this is a new problem, and that a new problem needs a new proof. It does not, and the reason is worth more than the answer. If you can turn a new problem into an old one in disguise, you do not need a new proof at all. You need a change of name.
Here is the change of name. Write d for b plus c. One letter standing in for two, and nothing else has happened. Because now the expression reads a plus d, all squared. And that is the thing you already know how to do. a squared, plus two a d, plus d squared. Three terms, not six. Nothing new has been assumed. The old identity was applied exactly as it stands, to a and to d.
Now put b plus c back where d was. a squared has no d in it, so it does not move. Two a d becomes two a, times b plus c, which opens out into two a b plus two a c. And d squared becomes b plus c, all squared. Look hard at that last piece, because it is the point of the method. It is the two-letter identity again. The same one.
Nothing new is borrowed to finish. It opens into b squared, plus two b c, plus c squared. Collect everything, and there are six terms. a squared, b squared, c squared. Then two a b, two b c, two c a. Three squares, one for each letter. Three doubled products, one for each pair of letters. Written in that order it sticks. The squares first, then the pairs going round the triangle, a to b to c and back again.
And one thing that trips people up. Two a c and two c a are not two different terms. They are the same term the other way round. There is no seventh term to go looking for. Now the drawing, because the algebra gives the answer and the picture gives the reason. Take a square of side a plus b plus c. Cut it twice across and twice down, at the same places both ways.
Nine pieces. Three of them sit on the diagonal, and those three are squares. a by a, b by b, c by c. The other six are rectangles, and they come in three matching pairs. Two of a by b, two of b by c, two of c by a. That drawing was rebuilt on one thousand seven hundred and twenty eight sets of lengths, and the nine pieces filled the square every time.
Each letter meets itself once, on the diagonal. That is where the three squares come from. Each pair of letters meets twice. Once in this row and that column, once the other way about. That is where the doubling comes from. Which kills a very common guess. There is not one cross term per letter. There is one per pair. With four letters you would get four squares and six doubled products, not four of each.
Two letters give three terms in all, three give six, four give ten, and five give fifteen. Time to make it earn its keep. Square a hundred and nineteen. Split it as a hundred, plus ten, plus nine, and read the six pieces straight off. The three squares: ten thousand, a hundred, eighty one. The three doubled products: two thousand, one thousand eight hundred, and a hundred and eighty. Add them. Fourteen thousand, one hundred and sixty one.
Every three-digit number splits like that. Tried on nine hundred and ninety of them, right every time. Which raises a question about judgement rather than correctness. Seventy eight is eighty minus two. Two parts, three terms, and you are finished. Six thousand and eighty four. A hundred and ninety eight is two hundred minus two. Thirty nine thousand, two hundred and four. Two hundred and fourteen has no round number near it, so it wants the three-way split. Forty five thousand, seven hundred and ninety six.
But a hundred and seventeen can be done either way. A hundred plus ten plus seven gives six pieces to write down. A hundred and twenty, minus three, gives three. Both come to thirteen thousand, six hundred and eighty nine. One of them is half the writing. That is the entire question being asked. Two things about minus signs. The first is that a minus needs no new identity at all. Replace b by minus b, and the same six terms come out with exactly two signs flipped.
The two that flipped are the two containing b. Two a b and two b c go negative. Two c a does not, because b was never in it. The second is a warning. Here is an expression that looks like a three-letter square and is not one. Four x squared, plus nine y squared, plus thirty six z squared, plus twenty four x y, plus thirty six y z, plus twelve z x.
The three squares name two x, three y and six z. The three cross terms want two x, six y and three z. Those disagree, so it is not a square. And here is the sting. Swap the nine and the thirty six, and it becomes two x plus six y plus three z, squared. Swap the twenty four and the twelve instead, and it becomes two x plus three y plus six z. Two repairs, two different answers.
Now a different use for an identity you have had for years. a squared minus b squared is a plus b, times a minus b. Move the b squared across the equals sign and it says something else entirely. a squared is a plus b, times a minus b, plus b squared. Same fact, different job: one factorises a difference, the other computes a square. And you can watch it happen. Take a square of side a. Cut a strip of height b off the bottom, and cut that strip at width a minus b.
Take the long piece, turn it through a right angle, and lay it against the right hand side. It keeps its area - it was a minus b across and b up, and now it is b across and a minus b up. What is left standing is one rectangle, a plus b across and a minus b up, and one small square of side b in the corner, which never moved. Checked on seven hundred and eighty pairs of lengths.
That is worth real money on numbers ending in five. Fifty five squared. Take b to be five. Sixty times fifty, plus twenty five. Three thousand and twenty five, and nothing awkward was ever multiplied. Thirty five squared is forty times thirty, plus twenty five. One thousand two hundred and twenty five. Sixty five gives four thousand two hundred and twenty five. Eighty five gives seven thousand two hundred and twenty five. A hundred and five gives eleven thousand and twenty five.
Now look only at the fronts. Twelve, forty two, seventy two, a hundred and ten. Three fours. Six sevens. Eight nines. Ten elevens. Any number ending in five is ten n plus five, and its square is n times n plus one, in hundreds, followed by twenty five. Tested on four hundred of them. One last thing, and it is a warning about trusting a pattern. Somebody offers you three squares. a plus b minus c, a minus b plus c, and a minus b minus c - and claims they add to twice a squared, plus twice b squared, plus twice c squared.
Test it. Let a and b and c all be one. The left side is one plus one plus one, which is three. The right side is six. It is not an identity. Expand it and the cross terms are still sitting there. They have not cancelled. It comes out right on thirty one triples out of seven hundred and twenty nine. That is not a rule, that is a coincidence.
Now add the fourth one, a plus b plus c squared, and everything changes. Every cross term cancels, in fours. What is left is the square of two a, plus the square of two b, plus the square of two c. Three of those squares do not balance. Four of them do.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- An identity holds for every value; an equation need notClass 9 · Ch 4, Exploring Algebraic Identities
- Reading (a + b)² off a partitioned squareClass 9 · Ch 4, Exploring Algebraic Identities
Comes up again in
- Sum and difference of cubes, and the three-term cubic identityClass 9 · Ch 4, Exploring Algebraic Identities
- Simplifying a rational expression, and the factor you must not cancelClass 9 · Ch 4, Exploring Algebraic Identities
Either side of this one
- Recognising an expression as an identity in disguiseClass 9 · Ch 4, Exploring Algebraic Identities
- Algebra tiles: factorising by rebuilding the rectangleClass 9 · Ch 4, Exploring Algebraic Identities