Express each of the following product as a monomials and verify the result in each case for x = 1:
$\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)$
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 law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$.
We have:
$\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)$
$=\left\{4 \times(-3) \times \frac{4}{5}\right\} \times\left(x^{2} \times x \times x^{3}\right)$
$=\left\{4 \times(-3) \times \frac{4}{5}\right\} \times\left(x^{2+1+3}\right)$
$=-\frac{48}{5} x^{6}$
$\therefore\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)=-\frac{48}{5} x^{6}$
Substituting x = 1 in LHS, we get:
$\mathrm{LHS}=\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)$
$=\left(4 \times 1^{2}\right) \times(-3 \times 1) \times\left(\frac{4}{5} \times 1^{3}\right)$
$=4 \times(-3) \times \frac{4}{5}$
$=-\frac{48}{5}$
Putting x = 1 in RHS, we get:
RHS $=-\frac{48}{5} x^{6}$
$=-\frac{48}{5} \times 1^{6}$
$=-\frac{48}{5}$
$\because L H S=\mathrm{RHS}$ for $x=1$; therefore, the result is correct
Thus, the answer is $-\frac{48}{5} x^{6}$.