ПОЖАЛУЙСТА РЕШИТЕ !!!!!!! (343x^3+125y^3/7x-5y)*(49x^2-70xy+25y^2/49x^2-35xy+25y^2)...

Решите задачу:

$\frac{7^{3}x^{3}+5^{3}y^{3}}{7x-5y}* \frac{7^{2}x^{2}-7*5*2xy+5^{2}y^{2}}{49x^{2}-35xy+25y^{2}}= \frac{(7x+5y)(49x^{2}-35xy+25y^{2})}{7x-5y}*\frac{(7x-5y)^{2}}{49x^{2}-35xy+25y^{2}} \\ =(7x+5y)(7x-5y)=49x^{2}-25y^{2}$
$49* (\sqrt{31})^{2}-25(\sqrt{47})^{2}=49*31-25*47=1519-1175=344$