If 1-y1-2 x-4 x2-8 x3-16 x4-32 x5-1-y6- where y-1- y-1-2- then the value of y-x is

The correct option is C 2(1 y)(1+2x+4x^2+8x^3+16x^4+32x^5) = 1 y^6 ⇒ (1 y)(1((2x)^6 1)2x 1)=1 y^6 ⇒ nbsp;(2x)^6 12x 1 = y^6 1y 1 ⇒ y = 2x ∴yx = 2 ...

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If 1-y1+2 x+4...

Question

If $(1 - y) (1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + 32 x^{5}) = 1 - y^{6},$ where $y \neq 1, x \neq \frac{1}{2}$ , then the value of $\frac{y}{x}$ is

\frac{1}{2}

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\frac{25}{24}

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\frac{24}{25}

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(1 - y) (1 + 2 x + 4 x^{2} + 8 x^{3} + 16 x^{4} + 32 x^{5}) = 1 - y^{6},

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\frac{y}{x}

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