Unpacking the Square

Take any three consecutive EVEN numbers, say 6, 8 and 10. Square the smallest and the largest, add those two squares, then subtract twice the middle square: 6^2 + 10^2 - 2(8^2) = 36 + 100 - 128 = 8. Now try 10, 12, 14: 10^2 + 14^2 - 2(12^2) = 100 + 196 - 288 = 8. Try any three consecutive even numbers, and the answer is always 8. This looks like magic until it is written algebraically. Calling the middle even number n, the three numbers are n-2, n, n+2. Then (n-2)^2 + (n+2)^2 - 2n^2 = (n^2 - 4n + 4) + (n^2 + 4n + 4) - 2n^2 = 8, with every n-term cancelling out. The 'magic' constant, 8, is really just four times the gap between consecutive terms (here the gap is 2, and 4 x 2^2 = ... in fact the constant equals twice the square of the gap: 2 x 2^2 = 8). An identity like this is not a coincidence spotted by luck, it is a fact that algebra can prove will hold for every possible choice of numbers fitting the pattern.

A square of side (a+b) can be split into a square of side a, a square of side b, and two rectangles of sides a and b, so its total area is a^2 + 2ab + b^2. This picture is not limited to positive lengths. Testing a = -5 and b = 2: (a+b)^2 = (-3)^2 = 9, while a^2 + 2ab + b^2 = 25 + 2(-5)(2) + 4 = 25 - 20 + 4 = 9. It matches, even though a is negative and cannot literally be drawn as a length. This confirms that (a+b)^2 = a^2 + 2ab + b^2 is an identity, true for every value of a and b, not just an equation true for a lucky few. Checking with a = 7, b = 4 also works: (11)^2 = 121, and 7^2 + 2(7)(4) + 4^2 = 49 + 56 + 16 = 121. What it is NOT equal to is a^2 + b^2: for a=7, b=4, that would wrongly give 49+16=65, missing the crucial cross term 2ab entirely.

Replacing b with -b in (a+b)^2 = a^2 + 2ab + b^2 gives (a-b)^2 = a^2 - 2ab + b^2. This shortcut can speed up mental squaring: to find 97^2, write 97 as (100-3), so 97^2 = 100^2 - 2(100)(3) + 3^2 = 10000 - 600 + 9 = 9409. What if three numbers are being squared together, (a+b+c)^2? Treating (b+c) as a single block d, and reusing (a+d)^2 = a^2+2ad+d^2, then expanding d^2 = (b+c)^2 = b^2+2bc+c^2, gives (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca. Checking this on 213, written as 200+10+3: 213^2 = 200^2+10^2+3^2+2(200)(10)+2(200)(3)+2(10)(3) = 40000+100+9+4000+1200+60 = 45369, which matches 213 x 213 exactly. Both of these are really the same trick as (a+b)^2, just applied with a negative term, or with three terms instead of two.

Every identity that expands a bracket also works in reverse, to factor an expression back into brackets. The difference-of-squares identity, a^2 - b^2 = (a+b)(a-b), instantly factors 64p^2 - 25q^2 by spotting that 64p^2 = (8p)^2 and 25q^2 = (5q)^2, giving 64p^2 - 25q^2 = (8p+5q)(8p-5q). A different, equally useful identity is (x+a)(x+b) = x^2+(a+b)x+ab, which factors quadratics like x^2+9x+20 by finding two numbers that add to 9 and multiply to 20. Since 4+5=9 and 4x5=20, x^2+9x+20 = (x+4)(x+5). The two numbers cannot be chosen separately for the sum and the product; they must satisfy both conditions together, which usually narrows the search to just one pair once the factors of the constant term are listed out.

Just as a square of side (a+b) splits into pieces giving (a+b)^2, a cube of edge (a+b) splits into two smaller cubes and six cuboids, giving (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. Replacing b with -b gives (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3, with the four terms alternating in sign. These identities can be used in reverse too: is 8x^3 + 36x^2 + 54x + 27 a perfect cube? Comparing it with a^3+3a^2b+3ab^2+b^3, try a = 2x and b = 3: (2x)^3 = 8x^3, 3(2x)^2(3) = 36x^2, 3(2x)(3)^2 = 54x, and 3^3 = 27, all matching exactly. So 8x^3+36x^2+54x+27 = (2x+3)^3. This can be checked numerically: at x=1, the expression gives 8+36+54+27 = 125, and (2(1)+3)^3 = 5^3 = 125, confirming the factoring is correct.

Multiplying out (x+y+z)(x^2+y^2+z^2-xy-yz-zx) leads, after every cross term cancels except a chosen few, to a strikingly clean result: x^3+y^3+z^3-3xyz. This identity is useful whenever three numbers' sum, product and sum-of-squares are known, but their sum-of-cubes is wanted. Suppose three numbers have sum 12, product 20, and sum of squares 62. First, find xy+yz+zx using (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx): 144 = 62 + 2(xy+yz+zx), so xy+yz+zx = 41. Then x^2+y^2+z^2-xy-yz-zx = 62-41 = 21. Substituting into the identity: x^3+y^3+z^3-3xyz = (12)(21) = 252. Since 3xyz = 3(20) = 60, the sum of cubes is x^3+y^3+z^3 = 252+60 = 312. A quantity that looked out of reach, the sum of three cubes, was found without ever knowing the three numbers themselves.

A rectangular vegetable patch is 3 metres longer than it is wide, and has an area of 70 square metres. Calling the width x metres, the length is x+3, and the area condition gives x(x+3) = 70, or x^2+3x-70 = 0. To factor this, two numbers are needed that add to 3 and multiply to -70: trying pairs of factors of 70 with opposite signs, 10 and -7 work, since 10 + (-7) = 3 and 10 x (-7) = -70. So x^2+3x-70 = x^2+10x-7x-70 = x(x+10) - 7(x+10) = (x+10)(x-7) = 0. This means either x+10 = 0 or x-7 = 0, giving x = -10 or x = 7. Since x is a width, it cannot be negative, so x = -10 is rejected, and the width is 7 metres, with length 7+3 = 10 metres. Checking: 7 x 10 = 70, exactly as required. An identity that once only expanded brackets has, by working in reverse, just solved an equation of degree 2, a genuinely new kind of power for algebra.