Question

if t square - 4t + 1=0 then the value of t cube + 1/t cube is = ?
solv plzz

Answer 1

=> t² – 4t + 1 =0

=> t² + 1 = 4t

=> ( t² + 1 ) ÷ t = 4

$=> \frac{ {t}^{2} }{t} + \frac{1}{t} = 4 \\ \\ => t + \frac{1}{t} = 4$




Now, Cubing on both sides,


$= > {(t + \frac{1}{t}) }^{3} = {4 }^{3} \\ \\ \\ => {t}^{3} + \frac{1}{t ^{3} } + 3(t + \frac{1}{t} )(t \times \frac{1}{t} ) = 64 \\ \\ \\ => {t}^{3} + \frac{1}{ {t}^{3} } + 3(4)(1) = 64 \\ \\ \\ => {t}^{3} + \frac{1}{ {t}^{3} } = 64 - 12 \\ \\ \\ = > {t}^{3} + \frac{1}{ {t}^{3} } = 52$

Answer 2

Answer:

Value of $t^3+\frac{1}{t^3}$ is 52.

Step-by-step explanation:

Given: t² - 4t + 1 = 0

To find: Value of $t^3+\frac{1}{t^3}$

Using quadratic formula we find the solution of given quadratic equation.

$t=\frac{-b\pm\sqrt{b^-4ac}}{2a}=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)}$

$t=\frac{4\pm\sqrt{16-4}}{2}=\frac{4\pm2\sqrt{3}}{2}=2\pm\sqrt{3}$

let, t = 2 + √3

then,

$\frac{1}{t}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=\frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2}$

$t=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

⇒ t + 1/t = 2 + √3 + 2 - √3 = 2 + 2 = 4

So,

(a + b)³ = a³ + b³ + 3ab( a + b )

a³ + b³ =(a + b)³ - 3ab( a + b )

put, a = t and b = 1/t

we get

$t^3+\frac{1}{t^3}=(t+\frac{1}{t})^3-3(t)(\frac{1}{t})(t+\frac{1}{t})=(4)^3-3(4)=64-12=52$

Therefore, Value of $t^3+\frac{1}{t^3}$ is 52.