## College Algebra 7th Edition

$x=-\displaystyle \frac{5}{3}$ $x=-\displaystyle \frac{4}{3}$
$(3x+5)^{4}-(3x+5)^{3} =0$ We can substitute $y$ for the repeating term: $y=3x+5$ The new equation is thus: $y^{4}-y^{3} =0$ We factor and solve: $y(y^{3}-1)=0$ $y(y-1)(y^{2}+y+1)=0$ $y=0$ or $y-1=0$ or $y^{2}+y+1=0$ $y=0$ or $y=1$ (The last equation $y^{2}+y+1=0$ has no real solutions because the discriminant is negative.) We need to solve for $x$ given the solutions for $y$: $y=3x+5$ $x=\frac{y-5}{3}$ If $y=0$, then $x=\frac{0-5}{3}=\frac{-5}{3}$ If $y=1$, then $x=\frac{1-5}{3}=\frac{-4}{3}$ So the solutions are: $x=-\displaystyle \frac{5}{3}$ and $x=-\displaystyle \frac{4}{3}$