Question : The value of $(x^{b+c})^{b–c}(x^{c+a})^{c–a}(x^{a+b})^{a–b}$, where $(x\neq 0)$ is:
Option 1: 1
Option 2: 2
Option 3: –1
Option 4: 0
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Correct Answer: 1
Solution : Given: $(x^{b+c})^{b–c}(x^{c+a})^{c–a}(x^{a+b})^{a–b}$ $(x\neq 0)$ = $(x^{(b+c)×(b–c)})(x^{(c+a)(c–a)})(x^{(a+b)×(a–b)})$ = $(x^{(b^{2}–c^{2})})(x^{(c^{2}–a^{2})})(x^{(a^{2}–b^{2})})$ = $x^{(b^{2}–c^{2}+c^{2}–a^{2}+a^{2}–b^{2})}$ = $x^{0}$ = 1 Hence, the correct answer is 1.
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Question : If $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$, then
Option 1: $a=b=c$
Option 2: $a\neq b=c$
Option 3: $a=b\neq c$
Option 4: $a\neq b\neq c$
Question : If $xy+yz+zx=0$, then $(\frac{1}{x^2–yz}+\frac{1}{y^2–zx}+\frac{1}{z^2–xy})$$(x,y,z \neq 0)$ is equal to:
Option 1: $3$
Option 2: $1$
Option 3: $x+y+z$
Option 4: $0$
Question : If (x – 8)2 – (x + 8)2 = 0. What is the value of x?
Option 2: 32
Option 3: 0
Option 4: –32
Question : If $x+\frac{1}{x}=0$, then the value of $x^{5}+\frac{1}{x^{5}}$ is:
Option 1: 2
Option 2: –1
Option 3: 1
Question : If $a+b+c=0$ and $a^2+b^2+c^2=40$, then what is the value of $a b+b c+c a$?
Option 1: –30
Option 2: –20
Option 3: –25
Option 4: –40
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