Question

if t + 1 divided by t = 8 then find the value of t cube +1/t cube​

Answer 1

.t + 1/t = 8

.(t + 1/t)^3 = t3 + 1/t3 + 3t + 3/t = t3 + 1/t3 + 3(t + 1/t)

. 8^3 = t3 + 1/t3 + 3×8

.512 = t3 + 1/t3 + 24

.488 = t3 + 1/t3

Hence ans. is 488

Hope it helps

Cheers

Answer 2

Step-by-step explanation:

Question:-

$\mathrm{If \: t + \frac{1}{t} = 8 \: then \: find \: the \: value \: of \: {t}^{3} + \frac{1}{ {t}^{3} } }$

Solution:-

Let's solve the problem

we have,

$\mathrm{t + \frac{1}{t} } = 8$

Cubing on both sides, using algebraic Identity:

(a+b)³ = a³ + b³ + 3ab (a+b)

we get

$\mathrm{ \bigg(t + \frac{1}{t} \bigg)^{3} = (8 {)}^{3} }$

$⟹ \mathrm{ {t}^{3} + \frac{1}{ {t}^{3} } + 3( \cancel{t}) \bigg( \frac{1}{ \cancel{t}} \bigg) \bigg(t + \frac{1}{t} \bigg) = 512}$

$⟹ \mathrm{ {t}^{3} + \frac{1}{ {t}^{3} } + 3 \bigg(t + \frac{1}{t} \bigg) = 512 }$

$⟹ \mathrm{{t}^{3} + \frac{1}{ {t}^{3} } + 3(8) = 512}$

$⟹ \mathrm{{t}^{3} + \frac{1}{ {t}^{3} } + 24 = 512}$

$⟹ \mathrm{{t}^{3} + \frac{1}{ {t}^{3} } = 512 - 24}$

$⟹ \mathrm{{t}^{3} + \frac{1}{ {t}^{3} } = 488}$

Answer:-

$\mathrm{Hence \: the \: value \: of \: {t}^{3} + \frac{1}{ {t}^{3} } \: is \: 488. }$

Used formulae:-

(a+b)³ = a³+b³+3ab(a+b)

:)