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Expand $\left(\frac{x}{3}+\frac{y}{5}\right)^3$ |
$\frac{x^3}{27}+\frac{x^2 y}{25}+\frac{x y^2}{25}+\frac{y^3}{125}$ $\frac{x^3}{25}+\frac{x^2 y}{15}+\frac{x y^2}{25}+\frac{y^3}{125}$ $\frac{x^3}{27}+\frac{x y}{15}+\frac{x y^2}{25}+\frac{y^3}{125}$ $\frac{x^3}{27}+\frac{x^2 y}{15}+\frac{x y^2}{25}+\frac{y^3}{125}$ |
$\frac{x^3}{27}+\frac{x^2 y}{15}+\frac{x y^2}{25}+\frac{y^3}{125}$ |
(a + b)3 = a3 + b3 + 3ab(a+b) $\left(\frac{x}{3}+\frac{y}{5}\right)^3$ So by expanding this cubic equation we get, $\frac{x^3}{27}+\frac{x^2 y}{15}+\frac{x y^2}{25}+\frac{y^3}{125}$ |
