Express each of the following product as a monomials and verify the result in each case for x = 1:

Solution:

 We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ $a^{m} \times a^{n}=a^{m+n}$ and $\left(a^{m}\right)^{n}=a^{m n}$

We have:

$\left(x^{2}\right)^{3} \times(2 x) \times(-4 x) \times 5$

$=\left(x^{6}\right) \times(2 x) \times(-4 x) \times 5$

$=\{2 \times(-4) \times 5\} \times\left(x^{6} \times x \times x\right)$

$=\{2 \times(-4) \times 5\} \times\left(x^{6+1+1}\right)$

$=-40 x^{8}$

$\therefore\left(x^{2}\right)^{3} \times(2 x) \times(-4 x) \times 5=-40 x^{8}$

Substituting x = 1 in LHS, we get:

LHS $=\left(x^{2}\right)^{3} \times(2 x) \times(-4 x) \times 5$

$=\left(1^{2}\right)^{3} \times(2 \times 1) \times(-4 \times 1) \times 5$

$=1^{6} \times 2 \times(-4) \times 5$

$=1 \times 2 \times(-4) \times 5$

$=-40$

Putting x = 1 in RHS, we get:

$\mathrm{RHS}=-40 x^{8}$

$=-40(1)^{8}$

$=-40 \times 1$

$=-40$

$\because L H S=R H S$ for $x=1$; therefore, the result is correct

Thus, the answer is $-40 x^{8}$.