Twice the sum of squares

Start from what is already proved. The two standard identities, both derived earlier from the distributive law, are the only raw material needed here:

where: $a$ and $b$ are any two real numbers (or algebraic terms), and $2ab$ is the "cross term" that distinguishes the sum-square from the difference-square.

Add them. Because (i) has $+2ab$ and (ii) has $-2ab$, adding the two equations makes that term vanish, and the two $a^2$'s and two $b^2$'s combine:

where: the right side is read as "twice the sum of the squares of $a$ and $b$" — the phrase that names this whole topic.

Subtract them. Subtracting (ii) from (i) does the opposite: the $a^2$ and $b^2$ terms cancel (they are identical in both), and the cross terms combine instead of cancel, since $-(-2ab)=+2ab$:

Why this matters beyond the algebra. These two combined identities let you move between three different pairs of quantities — $\{a+b,\,a-b\}$, $\{a^2+b^2,\,ab\}$, and $\{a,\,b\}$ themselves — without ever expanding a bracket by hand. If a question gives you $a+b$ and $a-b$ but hides $a$ and $b$, you can still recover $a^2+b^2$ (halve the sum of the two squares) and $ab$ (quarter the difference of the two squares).

Why the cancellation is not a coincidence. Look again at where the $2ab$ term comes from: it appears in $(a+b)^2$ because the cross terms $ab+ba$ from multiplying out $(a+b)(a+b)$ point in the "same direction," and in $(a-b)^2$ because the cross terms $-ab-ba$ from $(a-b)(a-b)$ point in the "opposite direction." Adding two things pointing in opposite directions is exactly what makes them cancel — the algebra is simply restating, in symbols, the everyday idea that $+2ab$ and $-2ab$ undo each other. Seeing this reason, rather than just memorising the outcome, is what lets you predict in advance that adding will kill the cross term and subtracting will kill the squares, without having to expand anything to check.

Verify $(a+b)^2+(a-b)^2 = 2(a^2+b^2)$ for $a=6$, $b=4$, then use the identity in reverse to recover $a^2+b^2$ from only $a+b$ and $a-b$.

Step 1. Compute the left side directly: $a+b=10$ and $a-b=2$, so $(a+b)^2+(a-b)^2 = 10^2+2^2 = 100+4 = 104$.

Step 2. Compute the right side directly: $a^2=36$ and $b^2=16$, so $2(a^2+b^2) = 2(36+16) = 2(52) = 104$.

Step 3. Both sides equal $104$, confirming the identity for this pair of values.

Step 4. Now use the identity in reverse. Suppose only $a+b=10$ and $a-b=2$ were known, with $a$ and $b$ themselves hidden. Then $a^2+b^2 = \dfrac{(a+b)^2+(a-b)^2}{2} = \dfrac{100+4}{2} = 52$, which matches the direct value $36+16=52$. The identity holds, and $a^2+b^2=52$ can be found without ever computing $a$ or $b$ separately.

Competitive stretch

A square garden of side $(x+y)$ metres and a smaller square garden of side $(x-y)$ metres are built side by side. Show that their combined area equals twice the sum of an $x$-side square and a $y$-side square.

Step 1. Area of the larger garden is $(x+y)^2$, and area of the smaller garden is $(x-y)^2$, so the combined area is $(x+y)^2+(x-y)^2$.

Step 2. By the identity just derived, this equals $2(x^2+y^2)$ directly, with no need to expand either bracket.

Step 3. Interpret the result: $2(x^2+y^2)$ is exactly twice the combined area of a square of side $x$ (area $x^2$) and a square of side $y$ (area $y^2$). So the two gardens together always cover the same ground as two copies of an $x$-square plus a $y$-square, for any $x>y>0$ — the identity re-read geometrically, using the same cancellation applied to areas instead of bare numbers.

• Re-expanding from scratch. Since $(a+b)^2$ and $(a-b)^2$ are already known, adding them directly (letting $2ab$ cancel) is faster and less error-prone than expanding $(a+b)^2+(a-b)^2$ term by term from the original brackets.
• Mixing up "sum" and "difference" results. Adding gives $2(a^2+b^2)$, with no $ab$ term. Subtracting gives $4ab$, with no squared terms at all. Confusing which operation produces which result is a common slip.
• Losing a sign when subtracting. $(a-b)^2$ already contains $-2ab$, so subtracting the whole expression turns that term into $-(-2ab)=+2ab$; dropping the sign flip is a frequent error.
• Assuming order matters. Since both identities involve squaring, $2(a^2+b^2)$ is unchanged if $a$ and $b$ are swapped, even though $(a-b)^2\ne(b-a)$'s appearance might suggest otherwise — in fact $(a-b)^2=(b-a)^2$ because squaring removes the sign.
• Forgetting the reverse direction. Many questions do not ask you to verify the identity at all — they give $a+b$ and $a-b$ and ask for $a^2+b^2$ or $ab$ directly. Trying to first solve for $a$ and $b$ individually before squaring wastes time; halving or quartering the combined identities is faster.

Ask the tutor: if you would like to see how combining $(a+b)^2$ with $ab$ (instead of $(a-b)^2$) also gives $a^2+b^2$, or how this same combining habit extends to cube identities in later classes, ask the AI tutor to work through those with you step by step.