Question

If a⁴+1/a⁴=50 then find the value of a³+1/a³​

Answer 1

Answer:

Step-by-step explanation:

Given:-

$\rm a^4 + \dfrac{1}{a^4} = 50$

To find:-

$\rm a^3+ \dfrac{1}{a^3}$

Solution:-

$\rm a^4 + \dfrac{1}{a^4} = 50$

$\rm\dashrightarrow (a^2)^2 + \dfrac{1}{(a^2)^2} = 50$

We know that,

a²+b² = (a+b)² - 2ab

Hence,

$\rm\dashrightarrow \Bigg ( a^2+\dfrac{1}{a^2} \Bigg )^2 - 2 \times a^2\times\dfrac{1}{a^2}= 50$

$\rm\dashrightarrow \Bigg ( a^2+\dfrac{1}{a^2} \Bigg )^2 - 2= 50$

$\rm\dashrightarrow \Bigg ( a^2+\dfrac{1}{a^2} \Bigg )^2= 50+2$

$\rm\dashrightarrow \Bigg ( a^2+\dfrac{1}{a^2} \Bigg )^2= 52$

$\rm\dashrightarrow a^2+\dfrac{1}{a^2} = \sqrt{52}= 2\sqrt{13}\qquad\qquad ...[1]$

Now let's find the value of a+1/a

$\rm\dashrightarrow \Bigg ( a+\dfrac{1}{a} \Bigg )^2 = a^2 + \dfrac{1}{a^2} + 2$

From [1] , a²+1/a² = 2√13, ∴

$\rm \dashrightarrow\Bigg ( a+\dfrac{1}{a} \Bigg )^2 = 2\sqrt{13}+ 2$

$\rm\dashrightarrow a+\dfrac{1}{a} = \sqrt{ 2\sqrt{13}+ 2}$

$\rm\dashrightarrow a+\dfrac{1}{a} = \sqrt{ 2(\sqrt{13}+1)}$

$\rm\dashrightarrow a+\dfrac{1}{a} = 2(\sqrt{13}+1)^{\frac{1}{2}}$

Now, we will find the value of a³+1/a³

we know that,

a³+b³ = (a+b)³ - 3ab(a+b) , ∴

$\rm\dashrightarrow a^3+\dfrac{1}{a^3} = \Bigg ( a+\dfrac{1}{a}\Bigg )^3-3\times a \times \dfrac{1}{a} \Bigg ( a + \dfrac{1}{a}\Bigg )$

$\rm\dashrightarrow a^3+\dfrac{1}{a^3} = \Bigg ( a+\dfrac{1}{a}\Bigg )^3-3\Bigg ( a + \dfrac{1}{a}\Bigg )$

From [2]

$\rm\dashrightarrow a^3+\dfrac{1}{a^3} = \bigg ( 2(\sqrt{13}+1)^{\frac{1}{2}} \bigg )^3-3(2(\sqrt{13}+1)^\frac{1}{2} )$

$\rm\dashrightarrow a^3+\dfrac{1}{a^3} =2(\sqrt{13}+1)^{\frac{3}{2}}-3(2(\sqrt{13}+1)^\frac{1}{2} )$

Hence, solved.