Find the Value of \(27x^3+8y^3\), if \(3x+2y=20\) and \(xy=\frac{14}{9}\)
Solution:
Using identity:
\[ a^3+b^3=(a+b)^3-3ab(a+b) \]
Here,
\[ a=3x,\quad b=2y \]
\[ a+b=20 \]
\[ ab=(3x)(2y)=6xy =6\left(\frac{14}{9}\right) =\frac{28}{3} \]
\[ 27x^3+8y^3 = (20)^3-3\left(\frac{28}{3}\right)(20) \]
\[ = 8000-560 \]
\[ =7440 \]