Algebra

The five working identities

$(a+b)^2 = a^2 + 2ab + b^2 \quad (a-b)^2 = a^2 - 2ab + b^2 \quad a^2-b^2 = (a+b)(a-b)$ $a^3+b^3 = (a+b)(a^2-ab+b^2) \qquad a^3-b^3 = (a-b)(a^2+ab+b^2)$ Everything else in this subtopic is these five rearranged. Read them BOTH ways: expanding, and factoring a big expression into something you can divide out.

The symmetric machine: given $a \pm b$ and $ab$

Almost every question hands you $a+b$ (or $a-b$) and $ab$, then asks for a higher symmetric expression. Never solve for $a$ and $b$ — rebuild from the top: $a^2+b^2 = (a+b)^2 - 2ab \qquad a^3+b^3 = (a+b)^3 - 3ab(a+b) \qquad a^3-b^3 = (a-b)^3 + 3ab(a-b)$ $a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2$ With $a+b = 5$, $ab = 6$: $a^2+b^2 = 13$, then $a^4+b^4 = 169 - 2\cdot36 = 97$. Two identities chained — that is the whole game. For a minus-stem, $a^2+b^2$ works the same way from $(a-b)^2 + 2ab$.

Big numbers: match the identity first

Large decimals and cubes are always an identity in disguise. Scan the denominator's middle term:

• $p^2 - pq + q^2$ under a numerator $p^3+q^3$ → the fraction collapses to $p+q$ (since $p^3+q^3 = (p+q)(p^2-pq+q^2)$).
• $p^2 + pq + q^2$ with numerator $p^3-q^3$ → collapses to $p-q$. So $\frac{4.7^3 + 2.3^3}{4.7^2 - 4.7\times2.3 + 2.3^2} = 4.7 + 2.3 = 7$ in one line. For sums like $873^2 + 127^2 + 873\times127$, notice the structure $a^2+b^2+ab = (a+b)^2 - ab$ with $a+b = 1000$.

Sum of squares = zero

A quadratic-type equation in several variables with real solutions is often a perfect-square sum: $x^2 + y^2 - 8x + 6y + 25 = (x-4)^2 + (y+3)^2 = 0$ Squares are never negative, so each must be exactly zero → $x = 4$, $y = -3$. The same trick with three variables: complete every square and force each to zero; then any requested combination ($x-y$, $x+y+z$, …) is read off directly. A close cousin: $a^2+b^2+c^2 = ab+bc+ca \iff a=b=c$, because the difference is half of $(a-b)^2+(b-c)^2+(c-a)^2$.

Direct evaluation

$(103\times97) = (100+3)(100-3) = 9991$ and $105^2 - 95^2 = 200\times10 = 2000$ — spot the $(a\pm b)$ forms around a round base. For products of two-digit numbers near 100, the base-100 split is faster than multiplication.

Common traps

• Expanding $(a+b)^3$ and forgetting the $3ab(a+b)$ term — always cross-check with a small case.
• Using $a^3+b^3$ with the WRONG trinomial ($p^2+pq+q^2$ belongs to $p^3-q^3$, not $p^3+q^3$).
• Trying to find $a$ and $b$ individually from $a+b$ and $ab$ — wasted time, and the surds get ugly.
• Missing that "sum of squares = 0" pins every variable exactly.

Quick revision

• $a^2+b^2 = (a\pm b)^2 \mp 2ab$; $a^3\pm b^3 = (a\pm b)^3 \mp 3ab(a\pm b)$; $a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2$.
• Denominator $p^2 \mp pq + q^2$ → fraction $= p \pm q$.
• $a^2+b^2+c^2 = ab+bc+ca \Rightarrow a=b=c$.
• Squares summing to zero → each square is 0.