Let $\alpha,\beta,\gamma$ and $\delta$ be the coefficients of $x^7,x^5,x^3$ and $x$ respectively in the expansion of $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$, $x1$. If $u$ and $v$ satisfy the equations $\alpha u+\beta v=18$, $\gamma u+\delta v=20$, then $u+v$ equals :

Question

Let $\alpha,\beta,\gamma$ and $\delta$ be the coefficients of $x^7,x^5,x^3$ and $x$ respectively in the expansion of $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$, $x>1$. If $u$ and $v$ satisfy the equations $\alpha u+\beta v=18$, $\gamma u+\delta v=20$, then $u+v$ equals :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The sum equals $2\left[x^5+10x^3(x^3-1)+5x(x^3-1)^2\right]=2\left[5x^7+10x^6+x^5-10x^4-10x^3+5x\right]$. Hence $\alpha=10,\ \beta=2,\ \gamma=-20,\ \delta=10$. The equations $10u+2v=18$ and $-20u+10v=20$ give $5u+v=9$ and $-2u+v=2$, so $u=1,\ v=4$ and $u+v=5$. Answer (1).

Q. Let $\alpha,\beta,\gamma$ and $\delta$ be the coefficients of $x^7,x^5,x^3$ and $x$ respectively in the expansion of $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$, $x>1$. If $u$ and $v$ satisfy the equations $\alpha u+\beta v=18$, $\gamma u+\delta v=20$, then $u+v$ equals :

Explanation

Reviewed by Dr. Abubacker

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