Question

if t^2-4t+1=0 then the value of t^3+1/t^3 is

Answer 1

See the attachment for detail solution
Hope it will help you

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.