Question : If $\frac{a}{b}+\frac{b}{a}=1$, then the value of $a^{3}+b^{3}$ will be:
Option 1: 1
Option 2: 0
Option 3: –1
Option 4: 2
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Correct Answer: 0
Solution : Given: $\frac{a}{b}+\frac{b}{a}=1$ ⇒ $\frac{a^2 +b^2}{ab}=1$ ⇒ ${(a^2 +b^2)}={ab}$__________(equation 1) Now, $a^{3}+b^{3}={(a+b)(a^2+b^2–ab)}$ Putting the value of ${a^2+b^2}$ from equation 1, we get: ${(a+b)(a^2+b^2–ab)}$ = ${(a+b)(ab–ab)}$ = ${(a+b)×0}=0$ Thus, $a^{3}+b^{3}=0$ Hence, the correct answer is $0$.
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Question : If $(a+\frac{1}{a})=–2$, then the value of $a^{1000}+a^{–1000}$ is:
Option 1: $2$
Option 2: $0$
Option 3: $1$
Option 4: $\frac{1}{2}$
Question : If $\text{a}= {\sqrt{2}+1}$ and $\text{b} = {\sqrt{2}–1}$, then the value of $\frac{1}{a+1}+\frac{1}{b+1}$ will be:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –1
Question : If $a+\frac{1}{a}=1$, then the value of $\frac{a^2-a+1}{a^2+a+1}$ is $(a\neq 0)$:
Option 2: –1
Option 3: 0
Question : If $\frac{a}{b}+\frac{b}{a}=-1$ and $a-b=2$, then the value of $a^3-b^3$ is:
Option 1: $0$
Option 2: $\frac{1}{2}$
Option 4: $–1$
Question : If $a+\frac{1}{a-2}=4$, then the value of $(a-2)^{2}+(\frac{1}{a-2})^{2}$ is:
Option 2: 2
Option 3: –2
Option 4: 4
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