Question 113
If $$2x+3y=\frac{11}{2}$$ and $$xy=\frac{5}{6}$$ then the value of $$8x^{3}+27y^{3}$$ is
Solution
$$2x+3y=\frac{11}{2}$$
cubing on both sides
$$(2x+3y)^{3}=(\frac{11}{2})^{3}$$
$$8x^{3}+27y^{3}+3(2x)(8y)(2x+3y)=\frac{1331}{8}$$
$$8x^{3}+27y^{3}+3(16xy)(2x+3y)=\frac{1331}{8}$$
$$8x^{3}+27y^{3}+3(16)(\frac{5}{6})(\frac{11}{2})=\frac{1331}{8}$$
$$8x^{3}+27y^{3}+220 =\frac{1331}{8}$$
$$8x^{3}+27y^{3} =\frac{1331}{8} - 220$$
$$8x^{3}+27y^{3} =\frac{1331}{8} - \frac{1760}{8} = - \frac{429}{8}$$
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