If t2-4t-1-0-Find t3-1t3-

Given that, t^2 4t + 1 = 0 t^2 #160; + 1 = 4t t #160; + 1t = 4 .......(i)Cubing on both sides, we get (t #160; + 1t)^3 = 4^3 t^3 ...

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Standard X

Mathematics

Quadratic Formula

If t2-4t+1=...

Question

If $t^{2} - 4 t + 1 = 0$ .
Find $t^{3} + \frac{1}{t^{3}}$ .

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Solution

Given that,
$t^{2} - 4 t + 1 = 0$

$⟹ t^{2} + 1 = 4 t$

$⟹ t + \frac{1}{t} = 4$ .......(i)
Cubing on both sides, we get

$⟹ {(t + \frac{1}{t})}^{3} = 4^{3}$

$⟹ t^{3} + \frac{1}{t^{3}} + 3. t . \frac{1}{t} (t + \frac{1}{t}) = 64$

$⟹ t^{3} + \frac{1}{t^{3}} + 3 (t + \frac{1}{t}) = 64$

$⟹ t^{3} + \frac{1}{t^{3}} + 3 (4) = 64$

$⟹ t^{3} + \frac{1}{t^{3}} = 64 - 12$

$∴ t^{3} + \frac{1}{t^{3}} = 52$

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If t2−4t+1=0.Find t3+1t3.

Given that,t2−4t+1=0⟹t2+1=4t⟹t+1t=4.......(i)Cubing on both sides, we get⟹(t+1t)3=43⟹t3+1t3+3.t.1t(t+1t)=64⟹t3+1t3+3(t+1t)=64⟹t3+1t3+3(4)=64⟹t3+1t3=64−12∴t3+1t3=52

If $t^{2} - 4 t + 1 = 0$ .
Find $t^{3} + \frac{1}{t^{3}}$ .