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# $(1-2y)\;(1+3x+9x^2+27x^3+81x^4+243x^5+729x^6)$ = $(1-64y^6)$, $(y\;\neq\;1)$, then $\frac{x}{y}$ is

$(a)\;\frac{3}{2}\qquad(b)\;\frac{2}{3}\qquad(c)\;\frac{4}{9}\qquad(d)\;\frac{9}{4}$

Answer : (b) $\;\frac{2}{3}$
Explanation : $\;(1-2y)\;(\frac{1-(3x)^6}{1-3x})=(1-64y^6)$
One possible value is $\;2y=3x$
$\frac{x}{y}=\frac{2}{3}.$