Question

Factor completely 4(x+1)^2/3 + 12(x+1)^-1/3

Answer 1

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad \sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\ \left.\qquad \qquad \right.\textit{negative exponents}\\\\ a^{-{ n}} \implies \cfrac{1}{a^{ n}} \qquad \qquad \cfrac{1}{a^{ n}}\implies a^{-{ n}} \qquad \qquad a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}} \\\\ -------------------------------\\\\ %4(x+1)^2/3 + 12(x+1)^-1/3 4(x+1)^{\frac{2}{3}}+12(x+1)^{-\frac{1}{3}}\implies 4(x+1)^{\frac{2}{3}}+12\cfrac{1}{(x+1)^{\frac{1}{3}}}$

$\bf 4(x+1)^{\frac{2}{3}}+\cfrac{12}{(x+1)^{\frac{1}{3}}}\impliedby \textit{so, our LCD is }(x+1)^{\frac{1}{3}} \\\\\\ \cfrac{4(x+1)^{\frac{2}{3}}\cdot (x+1)^{\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)^{\frac{2}{3}+\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}} \\\\\\ \cfrac{4(x+1)^{\frac{3}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4x+4+12}{(x+1)^{\frac{1}{3}}} \\\\\\ \cfrac{4x+16}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+4)}{\sqrt[3]{x+1}}$

Answer 2

4(x+1)^2/3  + 12(x+1)^-1/2

4(x+1)^2/3  + 12/(x+1)^1/3

[4(x+1) + 12 ]/(x+1)^1/3

(4x+16) / (x+1)^1/3