Question

Factor this expression. With work please!

Answer 1

$\bf \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\ \quad \\ \quad \\ % difference of cubes \textit{difference of cubes} \\ \quad \\ a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad (a+b)(a^2-ab+b^2)= a^3+b^3 \\ \quad \\ a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad (a-b)(a^2+ab+b^2)= a^3-b^3\\\\ -------------------------------$

$\bf x^6-9x^4-81x^2+729\qquad \begin{cases} x^6=x^{2\cdot 3}\\ \qquad (x^2)^3\\ 729=9\cdot 9\cdot 9\\ \qquad 9^3\\ 81=9\cdot 9\\ \qquad 9^2 \end{cases}\\\\ -------------------------------$

$\bf (x^6+729)-(9x^4+81x^2)\implies [(x^2)^3+9^3]-(9x^2x^2+9^2x^2) \\\\\\\ [\underline{(x^2+9)}(x^4-9x^2+9^2)]\ -\ [9x^2\underline{(x^2+9)}]\impliedby \begin{array}{llll} notice\ the\\ \textit{\underline{common factor}} \end{array} \\\\\\ (x^2+9)\ [(x^4\underline{-9x^2}+9^2)\underline{-9x^2}]\impliedby \textit{now we add \underline{these two}}$

$\bf (x^2+9)\ [x^4-18x^2+9^2]\impliedby \begin{cases} \sqrt{x^4}=x^2\\ \sqrt{9^2}=9\\ 18x^2=2(x^2)(9)\\ thus\ a\\ \textit{perfect square trinomial} \end{cases} \\\\\\ (x^2+9)\ [(x^2)^2-18x^2+9^2] \\\\\\ (x^2+9)(x^2-9)^2\implies (x^2+9)(x^2-3^2)^2 \\\\\\ (x^2+9)\ [(x-3)(x+3)]^2\implies \boxed{(x^2+9)(x-3)^2(x+3)^2} \\\\\\ \textit{and I guess you could be redundant and use} \\\\\\ (x^2+9)(x-3)(x-3)(x+3)(x+3)$