The expression [x + (x^3 - 1)^{1/2}]^5 + [x - | vedclass

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(C) Let $y = (x^3 - 1)^{1/2}$. The expression is $(x + y)^5 + (x - y)^5$.Using the binomial expansion,$(x + y)^5 + (x - y)^5 = 2[x^5 + ^5C_2 x^3 y^2 +

Vedclass · Accepted Answer

$7$

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(C) Let $y = (x^3 - 1)^{1/2}$. The expression is $(x + y)^5 + (x - y)^5$.Using the binomial expansion,$(x + y)^5 + (x - y)^5 = 2[x^5 + ^5C_2 x^3 y^2 + ^5C_4 x y^4]$.Substituting $y^2 = x^3 - 1$ and $y^4 = (x^3 - 1)^2 = x^6 - 2x^3 + 1$:$= 2[x^5 + 10x^3(x^3 - 1) + 5x(x^6 - 2x^3 + 1)]$$= 2[x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x]$$= 10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x$.The highest power of $x$ in the resulting polynomial is $7$,so the degree is $7$.

The expression $[x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5$ is a polynomial of degree

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