## College Algebra (10th Edition)

$\displaystyle \frac{4x+3}{3(x+1)^{2/3}}$
$(x+1)^{1/3}+x\displaystyle \cdot\frac{1}{3}(x+1)^{-2/3}=$ ... rewrite the second term moving $(x+1)$ into the denominator $=(x+1)^{1/3}+\displaystyle \frac{x}{3(x+1)^{2/3}}$ ... rewrite, with the first term being a fraction with the common denominator $=\displaystyle \frac{(x+1)^{1/3}\cdot 3(x+1)^{2/3}+x}{3(x+1)^{2/3}}$ ... apply $a^{m}\cdot a^{n}=a^{m+n}$ to the first term of the numerator $=\displaystyle \frac{3(x+1)^{1/3+2/3}+x}{3(x+1)^{2/3}}$ $=\displaystyle \frac{3(x+1)^{1}+x}{3(x+1)^{2/3}}$ $=\displaystyle \frac{3x+3+x}{3(x+1)^{2/3}}$ $=\displaystyle \frac{4x+3}{3(x+1)^{2/3}}$