Question

${t}^{3} + {(1 \div t)}^{3} = 52$
then find
${t}^{2} + {(1 \div t)}^{2}$
​

Answer 1

Answer:

$\large{\underline{\boxed{\sf t^2 + \dfrac{1}{t^2}= 14 }}}$

Step-by-step explanation:

Given that ;

$\sf t ^{3} + \frac{1}{t^{3} } = 52 \\ \\ \sf (t + \frac{1}{t})^{3} - 3 \: . \: t \: . \: \frac{1}{t} (t + \frac{1}{t} ) = 52 \\ \\ \sf (t + \frac{1}{t})^{3} -3(t + \frac{1}{t} ) = 52$

Let (t + $\sf \dfrac{1}{t}$ ) = x,

$\sf x {}^{3} - 3x = 52 \\ \\ \sf x^{3} - 3x - 52 = 0 \\ \\ \sf x {}^{3} - 4x {}^{2} + 4x {}^{2} - 16x + 13x - 52 = 0 \\ \\ \sf x^{2} (x - 4) + 4x(x - 4) + 13(x - 4) = 0 \\ \\ \sf (x - 4)( {x}^{2} + 4x + 13) = 0$

Here, we get

x = 4 or x² + 4x + 13 = 0

On solving the second equation ;

x² + 4x + 13 = 0

On comparing the given equation with ax² + bx + c = 0, so

• a = 1
• b = 4
• c = 13

Discriminant = b² - 4ac

⇒ D = (4)² - 4 * 1 * 13

⇒ D = 16 - 52

⇒ D = - 36

∴ D < 0

So, no real roots exist.

_______________________

Here, we got the value of x = 4.

$\sf (t + \frac{1}{t}) = 4$

On squaring both sides ;

$\sf (t + \frac{1}{t})^{2} = (4) {}^{2} \\ \\ \implies \sf {t}^{2} + \frac{1}{ {t}^{2} } + 2 \: . \:t \: . \: \frac{1}{t} = 16 \\ \\ \implies \sf {t}^{2} + \frac{1}{ {t}^{2} } + 2 = 16 \\ \\ \implies \sf {t}^{2} + \frac{1}{ {t}^{2} } = 16 - 2 \\ \\ \boxed{ \bf \therefore \: {t}^{2} + \frac{1}{ {t}^{2} } = 14}$

Answer 2

Solution :-

As given

$t^3 + \dfrac{1}{t^3} = 52$

Now as we know :-

a³ + b³ = (a + b)³ - 3ab(a + b) ....(i)

a² + b² = (a + b)² - 2ab .....(ii)

By using our first equation

$t^3 + \dfrac{1}{t^3} = \left( t + \dfrac{1}{t} \right)^3 - 3\times t \times \dfrac{1}{t} \left( t + \dfrac{1}{t} \right) = 52$

$\left( t + \dfrac{1}{t} \right)^3 - 3\left( t + \dfrac{1}{t} \right) = 52$

Now let

$\left( t + \dfrac{1}{t} \right) = y$

Then

$y^3 - 3y = 52$

$\implies y^3 - 3y - 52 = 0$

Now we will put y = 4 (via hit and trial)

$(4)^3 - 3(4) - 52 = 0$

$\implies 64 - 12 - 52 = 0$

$\implies 64 - 64 = 0$

Now as 4 is a factor

→ (y-4) is a factor of y³ - 3y - 52

Now via dividing it by (y - 4) we get

y² + 4y + 13 as a quotient

Now as we can clearly see that

D = 4² - 4 × 13

D = 16 - 52

D = - 36

So discriminant is negative hence it will not give any real roots.

Hence y = 4

$\implies \left( t + \dfrac{1}{t} \right) = 4$

Now the Value of

$\implies t^2 + \dfrac{1}{t^2} = \left( t + \dfrac{1}{t} \right)^2 - 2 \times t \times \dfrac{1}{t}$

$\implies t^2 + \dfrac{1}{t^2} = (4)^2 - 2$

$\implies t^2 + \dfrac{1}{t^2} = 16 - 2$

$\implies t^2 + \dfrac{1}{t^2} = 14$