Question

(5a²)·(-2a²b)³

Explanation:

Step1: Expand the power of the second term

First, we expand \((-2a^{2}b)^{3}\) using the power - of - a - product rule \((xy)^n=x^n y^n\) and \((x^m)^n = x^{mn}\).
\((-2a^{2}b)^{3}=(-2)^{3}\times(a^{2})^{3}\times b^{3}\)
We know that \((-2)^{3}=-8\), \((a^{2})^{3}=a^{2\times3}=a^{6}\), so \((-2a^{2}b)^{3}=-8a^{6}b^{3}\)

Step2: Multiply the two monomials

Now we multiply \((5a^{2})\) with \(-8a^{6}b^{3}\) using the rule of multiplying monomials \(x^m\times x^n=x^{m + n}\) (for the same base \(x\)) and the coefficient multiplication rule.
\((5a^{2})\cdot(-8a^{6}b^{3})=(5\times(-8))\times(a^{2}\times a^{6})\times b^{3}\)
The product of the coefficients \(5\times(-8)=-40\), and for the powers of \(a\), \(a^{2}\times a^{6}=a^{2 + 6}=a^{8}\)

So \((5a^{2})\cdot(-2a^{2}b)^{3}=-40a^{8}b^{3}\)

Answer:

\(-40a^{8}b^{3}\)