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Let $a, b, c$ and $d$ be any four real numbers. Then. $a^{n}+b^{n}=c^{n}+d^{n}$ holds for any natural number $n$. if
Options:
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Verified Answer
The correct answer is:
$a-b=c-d \cdot d^{2}-b^{2}=c^{2}-d^{2}$
From option (d), we have
$\begin{aligned}
a-b &=c-d \\
a^{2}-b^{2} &=c^{2}-d^{2}
\end{aligned}$
Consider, $a^{2}-b^{2}=c^{2}-d^{2}$
$\Rightarrow(a+b)(a-b)=(c-d)(c+d)$
$\Rightarrow \quad a+b=c+d$
On adding Eqs we get
$2 a=2 c \Rightarrow a=c$
$\Rightarrow \quad b=d$
Thus, $a^{n}+b^{n}=c^{n}+d^{n}$ for all $n \in N$
$\begin{aligned}
a-b &=c-d \\
a^{2}-b^{2} &=c^{2}-d^{2}
\end{aligned}$
Consider, $a^{2}-b^{2}=c^{2}-d^{2}$
$\Rightarrow(a+b)(a-b)=(c-d)(c+d)$
$\Rightarrow \quad a+b=c+d$
On adding Eqs we get
$2 a=2 c \Rightarrow a=c$
$\Rightarrow \quad b=d$
Thus, $a^{n}+b^{n}=c^{n}+d^{n}$ for all $n \in N$
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