$$-\frac{a}{5}\left(2\times\frac{a^2}{5}\right)^2+5\left(-\frac{a}{5}\right)^3\left(2\times\frac{a^2}{5}\right) -a^2\left(-\frac{a}{5}\right)\left(2\times\frac{a^2}{5}\right)=0$$
I get
$$\frac{-4a^5}{25}-\frac{2a^5}{125}+\frac{2a^5}{25}$$
Where I multiply the first and third fraction by 5 in the numerator and denominator to get 125 in the denominator
For some reason, I get the wrong answer of
$$-\frac{12a^5}{125}$$
When the answer should be
$$\frac{4a^5}{125}$$
What have I done wrong?
we have $$(2\frac{a^2}{5})\left(-\frac{2a^3}{25}-\frac{a^3}{25}+\frac{a^3}{5}\right)$$=$$\frac{2a^2}{5}\cdot \frac{2a^3}{25}=$$ $$\frac{4a^5}{125}=0$$ thus $$a=0$$