Question:
If \[ x=3 \quad \text{and} \quad y=-1 \] find the value of:
\[ \left(\frac{x}{4} – \frac{y}{3}\right) \left(\frac{x^2}{16} + \frac{xy}{12} + \frac{y^2}{9}\right) \]
Solution:
Using identity:
\[ (a-b)(a^2+ab+b^2)=a^3-b^3 \]
Here, \[ a=\frac{x}{4},\qquad b=\frac{y}{3} \]
\[ \left(\frac{x}{4} – \frac{y}{3}\right) \left(\frac{x^2}{16} + \frac{xy}{12} + \frac{y^2}{9}\right) \]
\[ = \left(\frac{x}{4}\right)^3 – \left(\frac{y}{3}\right)^3 \]
Substituting \[ x=3,\qquad y=-1 \]
\[ = \left(\frac{3}{4}\right)^3 – \left(\frac{-1}{3}\right)^3 \]
\[ = \frac{27}{64} + \frac{1}{27} \]
\[ = \frac{729+64}{1728} \]
\[ = \frac{793}{1728} \]