Question : Determine the value of $\small \left (\frac{1}{r}+\frac{1}{s} \right)$, when $r^{3}+s^{3}=0$ and $r+s=6$.
Option 1: 0
Option 2: 0.5
Option 3: 1
Option 4: 6
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Correct Answer: 0.5
Solution : We know that, $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$ So, we can write, $(r + s)^3 = r^3 + s^3 + 3rs(r + s)$ Substituting given values, we have, $6^3 = 0 + 3rs\times 6$ ⇒ $216 = 18rs$ ⇒ $rs = 12$ Now, $(\frac{1}r+ \frac{1}{s}) = \frac{(r + s)}{rs}$ ⇒ $(\frac{1}r+ \frac{1}{s}) = \frac{6}{12}$ = 0.5 Hence, the correct answer is 0.5.
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Question : If $\small x=a\left (b-c \right),\; y=b\left (c-a \right) ,\; z=c\left (a-b \right)$, then the value of $\left (\frac{x}{a} \right)^{3}+\left (\frac{y}{b} \right)^{3}+\left (\frac{z}{c} \right)^{3}$ is:
Option 1: $\frac{2xyz}{abc}$
Option 2: $\frac{xyz}{abc}$
Option 3: $0$
Option 4: $\frac{3xyz}{abc}$
Question : If $\small c+\frac{1}{c}=3$, then the value of $\left (c-3 \right )^{7}+\frac{1}{c^{7}}$ is:
Option 1: 2
Option 2: 0
Option 3: 3
Option 4: 1
Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Option 1: $\frac{2 \sqrt{6}-6}{5}$
Option 2: $\frac{4 \sqrt{6}-6}{5}$
Option 3: $\frac{4 \sqrt{6}-6}{3}$
Option 4: $\frac{4 \sqrt{3}-6}{5}$
Question : The value of $2 \frac{3}{5} \div\left[2 \frac{1}{3} \div\left\{4 \frac{1}{3}-\left(2 \frac{1}{2}+\frac{2}{3}\right)\right\}\right]$ is equal to:
Option 1: $1 \frac{3}{10}$
Option 2: $2 \frac{7}{10}$
Option 3: $2 \frac{3}{7}$
Option 4: $1 \frac{3}{7}$
Question : If $\left(3 y+\frac{3}{y}\right)=8$, then find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Option 1: $5\frac{1}{9}$
Option 2: $4\frac{5}{6}$
Option 3: $7\frac{1}{9}$
Option 4: $9\frac{1}{9}$
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