Write down the product of −

Solution:

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

We have:

$\left(-8 x^{2} y^{6}\right) \times(-20 x y)$

$=\{(-8) \times(-20)\} \times\left(x^{2} \times x\right) \times\left(y^{6} \times y\right)$

$=\{(-8) \times(-20)\} \times\left(x^{2+1}\right) \times\left(y^{6+1}\right)$

$=-160 x^{3} y^{7}$

$\therefore\left(-8 x^{2} y^{6}\right) \times(-20 x y)=-160 x^{3} y^{7}$

Substituting x = 2.5 and y = 1 in LHS, we get:

$\mathrm{LHS}=\left(-8 x^{2} y^{6}\right) \times(-20 x y)$

$=\left\{-8(2.5)^{2}(1)^{6}\right\} \times\{-20(2.5)(1)\}$

$=\{-8(6.25)(1)\} \times\{-20(2.5)(1)\}$

$=(-50) \times(-50)$

$=2500$

Substituting x = 2.5 and y = 1 in RHS, we get:​

$\mathrm{RHS}=-160 x^{3} y^{7}$

$=-160(2.5)^{3}(1)^{7}$

$=-160(15.625) \times 1$

$=-2500$

Because LHS is equal to RHS, the result is correct.

Thus, the answer is $-160 x^{3} y^{7}$.