Question : If $7 a-\frac{7}{a}+4=0$, then find $a^3-\frac{1}{a^3}-1$.
Option 1: $-\frac{995}{343}$
Option 2: $-\frac{875}{248}$
Option 3: $-\frac{694}{315}$
Option 4: $-\frac{765}{262}$
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Correct Answer: $-\frac{995}{343}$
Solution :
Given:
$7a-\frac{7}{a}+4=0$
⇒ $7(a-\frac{1}{a})=-4$
⇒ $(a-\frac{1}{a})=-\frac{4}{7}$
Cubing both sides, we get,
⇒ $a^3-\frac{1}{a^3}-3×a×\frac{1}{a}×(a-\frac{1}{a})=\frac{–64}{343}$
⇒ $a^3-\frac{1}{a^3}-3× (-\frac{4}{7})=-\frac{64}{343}$
⇒ $a^3-\frac{1}{a^3}=-\frac{64}{343}-\frac{12}{7}$
Adding –1 with both sides, we get,
⇒ $a^3-\frac{1}{a^3}-1=-\frac{64}{343}-\frac{12}{7}-1$
$\therefore a^3-\frac{1}{a^3}-1=-\frac{995}{343}$
Hence, the correct answer is $(\frac{–995}{343})$.
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