Question

if (a-1/a) = 4 , prove that a⁶-76a³-1=0​

Answer 1

=> (a - 1/a) = 4

Cube on both sides :

=> (a - 1/a)³ = 4³

=> (a)³ - (1/a)³ - 3(a)(1/a)(a - 1/a) = 64

=> a³ - 1/a³ - 3(1)(a - 1/a) = 64

=> a³ - 1/a³ - 3(1)(4) = 64

=> a³ - 1/a³ - 12 - 64 = 0

=> a³ - 1/a³ - 76 = 0

Multiply both sides by a³ or take LCM:

=> a³ × [a³ - 1/a³ - 76] = a³ × 0

=> a⁶ - 1 - 76a³ = 0

=> a⁶ - 76a³ - 1 = 0 proved

Formula used :

(a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

$\sf \leadsto ( a - \frac{1}{a} ) = 4 \\ \tt \: we \: know \: that \\ \sf \leadsto \boxed{ \sf \: (a - b) {}^{3} = {a}^{3} - b {}^{3} - 3ab(a - b)} \\ \sf \leadsto {(a - \frac{1}{a} )}^{3} = 4 {}^{3} \\ \sf \leadsto {a}^{ 3} - { \frac{1}{a} }^{3} - 3 \times a \times \frac{1}{a} (a - \frac{1}{a} ) = 64 \\ \sf \leadsto {a}^{3} - \frac{1}{a {}^{3} } - 3( \frac{ {a}^{2} - 1 }{a} ) = 64 \\ \sf \leadsto \frac{ {a}^{6} - 1 }{a} - \frac{3 {a}^{2} - 3}{a} = 64 \\ \sf \leadsto \frac{ {a}^{6} - 1 - 3 {a}^{2} + 3}{a} = 64 \\ \sf \leadsto \frac{a^6-1}{a}-3(4)=64 \\ \sf \leadsto a^6-1-12=64a \\ \sf \leadsto \frac{a^6-1}{a}=64+12=76 \\ \sf \leadsto a^6-1=76a \\ \sf \leadsto a^6-76a-1=0$

Hence proved