Question

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

Answer 1

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

Answer 2

`\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)^{(2)/(3)}+12(x+1)^{-(1)/(3)}\implies 4(x+1)^{(2)/(3)}+12\cfrac{1}{(x+1)^{(1)/(3)}} \\\\\\`


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