UPSKILL MATH PLUS

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Use the following identities to factorise the expressions.

$\begin{array}{l}{\left(a+b\right)}^{2}={a}^{2}+2\mathit{ab}+{b}^{2}\\ \\ {\left(a-b\right)}^{2}={a}^{2}-2\mathit{ab}+{b}^{2}\\ \\ {a}^{2}-{b}^{2}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\left(a+b\right)\left(a-b\right)\\ \\ {\left(a+b\right)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}\\ \\ {\left(a-b\right)}^{3}={a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}\\ \\ {a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-\mathit{ab}+{b}^{2}\right)\\ \\ {a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+\mathit{ab}+{b}^{2}\right)\end{array}$
Example:
1) $$y^3+64$$

Let us write the above expression as $$y^3+4^3$$.

Use the identity, $$a^3+b^3 = (a+b)(a^2-ab+b^2)$$.

${y}^{3}+{4}^{3}=\left(y+4\right)\left({y}^{2}-y\left(4\right)+{4}^{2}\right)$

${y}^{3}+{4}^{3}=\left(y+4\right)\left({y}^{2}-4y+16\right)$

2) $$9x^2 -16y^2$$

Let us write the above expression as $$(3x)^2-(4y)^2$$.

Use the identity, $$a^2-b^2$$ $$=$$ $$(a+b)(a-b)$$.

$$(3x)^2-(4x)^2$$ $$=$$ $$(3x+4y)(3x-4y)$$.