What is the value of $\sqrt[3]{2a^2 b^4} + \dfrac{a - c}{(b+c)^2}$ if $a = 4$,b = 2 c = -5?

ghistory Jun 17, 2024

#1**+1 **

We already have all the information we need to find \(\sqrt[3]{2a^2 b^4} + \dfrac{a - c}{(b+c)^2}\), so let's plug them in.

We get

\(\sqrt[3]{2(4)^2 (2)^4} + \dfrac{4 +5}{(2-5)^2}\)

Simplifying and multiplying everything out, we get

\(\sqrt[3]{512} + \dfrac{9}{(-3)^2}\\ 8+1\\ 9\)

So our final answer is 9.

I may have made a mistake with computation. If I did, let me know.

Thanks! :)

NotThatSmart Jun 17, 2024

#1**+1 **

Best Answer

We already have all the information we need to find \(\sqrt[3]{2a^2 b^4} + \dfrac{a - c}{(b+c)^2}\), so let's plug them in.

We get

\(\sqrt[3]{2(4)^2 (2)^4} + \dfrac{4 +5}{(2-5)^2}\)

Simplifying and multiplying everything out, we get

\(\sqrt[3]{512} + \dfrac{9}{(-3)^2}\\ 8+1\\ 9\)

So our final answer is 9.

I may have made a mistake with computation. If I did, let me know.

Thanks! :)

NotThatSmart Jun 17, 2024