Question 67

If $$x+3y+2=0$$ then value of $$x^{3}+27y^{3}+8-18xy$$ is:

Solution

$$x+3y+2=0$$

x + 3y = -2

Taking cube both sides,

$$(x + 3y)^3 = -8$$

$$x^3 + 27y^3 + 3x.3y(x + 3y) = -8$$

$$x^3 + 27y^3 + 9xy(-2) = -8 $$

$$x^{3}+27y^{3} -18xy = -8$$

$$x^{3}+27y^{3}+8-18xy$$ = 0

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