Question

If x = −2 and y = 1, by using an identity find the value of the following

$$(5y+\frac{15}{y})(25y^2-75+\frac{225}{y^2})$$

Answer 1

Given $$(5y+\frac{15}{y})(25y^2-75+\frac{225}{y^2})$$

We shall use the identity ${\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2)}$,

We can rearrange the  $$(5y+\frac{15}{y})(25y^2-75+\frac{225}{y^2})$$as

${\displaystyle =(5y+\frac{15}{y})[{\left(5y\right)}^2+{\left(\frac{15}{y}\right)}^2-\left(5y\right)\left(\frac{15}{y}\right)]}$

${\displaystyle ={\left(5y\right)}^3+{\left(\frac{15}{y}\right)}^3}$

 ${\displaystyle =\left(5y\right)\times \left(5y\right)\times \left(5y\right)+\left(\frac{15}{y}\right)\times \left(\frac{15}{y}\right)\times \left(\frac{15}{y}\right)}$

${\displaystyle =125y^3+\frac{3375}{y^3}}$

Now substituting the value  y = 1in  ${\displaystyle 125y^3+\frac{3375}{y^3}}$

${\displaystyle =125y^3+\frac{3375}{y^3}}$

 ${\displaystyle =125{\left(1\right)}^3+\frac{3375}{{\left(1\right)}^3}}$ 

${\displaystyle =125+3375}$ 

${\displaystyle =3500}$

Hence the Product value of  $$(5y+\frac{15}{y})(25y^2-75+\frac{225}{y^2})$$is  3500.