Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^7, x^5, x^3$ and $x$ respectively in the expansion of

$$\begin{aligned} & \left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5, x>1 \text {. If } u \text { and } v \text { satisfy the equations } \\\\ & \alpha u+\beta v=18, \\\\ & \gamma u+\delta v=20, \end{aligned}$$

then $\mathrm{u+v}$ equals :

<p>To find the sum of $ u $ and $ v $, we first need to expand the expression: </p> <p>$ \left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5 $</p> <p>Using the Binomial Theorem, the expansion yields:</p> <p>$ = 2\left({}^5C_0 \cdot x^5 + {}^5C_2 \cdot x^3(x^3-1) + {}^5C_4 \cdot x(x^3-1)^2\right) $</p> <p>Simplifying this, we obtain:</p> <p>$ = 2\left(5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x\right) $</p> <p>From this expansion, we can identify the coefficients:</p> <p><p>The coefficient of $ x^7 $ is $ \alpha = 10 $</p></p> <p><p>The coefficient of $ x^5 $ is $ \beta = 2 $</p></p> <p><p>The coefficient of $ x^3 $ is $ \gamma = -20 $</p></p> <p><p>The coefficient of $ x $ is $ \delta = 10 $</p></p> <p>Given the equations:</p> <p>$ \alpha u + \beta v = 18 $</p> <p>$ \gamma u + \delta v = 20 $</p> <p>Substituting in the coefficients:</p> <p>$ 10u + 2v = 18 $</p> <p>$ -20u + 10v = 20 $</p> <p>By solving these equations, we find:</p> <p><p>From $ 10u + 2v = 18 $, simplify to $ 5u + v = 9 $.</p></p> <p><p>From $ -20u + 10v = 20 $, simplify to $ -2u + v = 2 $.</p></p> <p>Solving these linear equations simultaneously, we find:</p> <p>Subtracting equation 2 from equation 1: </p> <p>$ 5u + v = 9 $</p> <p>$ - (-2u + v = 2) $</p> <p>This yields:</p> <p>$ 7u = 7 \quad \Rightarrow \quad u = 1 $</p> <p>Substitute $ u = 1 $ back into $ 5u + v = 9 $:</p> <p>$ 5(1) + v = 9 \quad \Rightarrow \quad v = 4 $</p> <p>Thus, the sum $ u + v $ is:</p> <p>$ u + v = 1 + 4 = 5 $</p>