Question:
If \[ x=3 \quad \text{and} \quad y=-1 \] find the value of:
\[ \left(\frac{x}{7} + \frac{y}{3}\right) \left(\frac{x^2}{49} + \frac{y^2}{9} – \frac{xy}{21}\right) \]
Solution:
Using identity:
\[ (a+b)(a^2-ab+b^2)=a^3+b^3 \]
Here, \[ a=\frac{x}{7},\qquad b=\frac{y}{3} \]
\[ \left(\frac{x}{7} + \frac{y}{3}\right) \left(\frac{x^2}{49} – \frac{xy}{21} + \frac{y^2}{9}\right) \]
\[ = \left(\frac{x}{7}\right)^3 + \left(\frac{y}{3}\right)^3 \]
Substituting \[ x=3,\qquad y=-1 \]
\[ = \left(\frac{3}{7}\right)^3 + \left(\frac{-1}{3}\right)^3 \]
\[ = \frac{27}{343} – \frac{1}{27} \]
\[ = \frac{729-343}{9261} \]
\[ = \frac{386}{9261} \]