Question

1. $-3a^{2}b cdot 7(a^{9})^{5}$

Explanation:

Step1: Simplify the power of a power

According to the power - of - a - power rule \((x^{m})^{n}=x^{m\times n}\), for \((a^{9})^{5}\), we have \(m = 9\) and \(n = 5\), so \((a^{9})^{5}=a^{9\times5}=a^{45}\)

Step2: Multiply the coefficients and the like - bases

First, multiply the coefficients: \(- 3\times7=-21\)
Then, multiply the terms with the base \(a\) according to the rule of multiplying powers with the same base \(x^{m}\cdot x^{n}=x^{m + n}\). For \(a^{2}\) and \(a^{45}\), we get \(a^{2+45}=a^{47}\)
The term with \(b\) remains unchanged.
So, \(-3a^{2}b\cdot7(a^{9})^{5}=-21a^{47}b\)

Answer:

\(-21a^{47}b\)