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If $2 x+3 y+1=0$, then what is the value of $\left(8 x^3+8+27 y^3-18 x y\right)$ ? |
-7 7 -9 9 |
7 |
We know that, (a + b)3 = a3 + b3 + 3ab(a + b) 2x + 3y + 1 = 0 = 2x + 3y = -1 Now, (2x + 3y)3 = (2x)3 + (3y)3 + 3 × 2x × 3y (-1) = -1 = 8x3 + 27y3 – 18xy = -1 So, the value of (8x3 + 8 + 27y3 – 18xy) = 8 – 1 = 7 |
