Question

find the value a+1/a = 4 then the value of a⁴-1/a⁴​

Answer 1

Step-by-step explanation:

a⁴-1/a⁴

=(a²)²-(1/a²)²

=(a²+1/a²)(a²-1/a²)

=(a²+1/a²)(a)²-(1/a)²

=(a²+1/a²)(a+1/a)(a-1/a)

Answer 2

➲CORRECT QUESTION :

• If the value of $\bf{a + \dfrac{1}{a} = 4 }$. Then, find the value of $\bf{ a^4 + \dfrac{1}{ a^4} }$

☯GIVEN :

• $\bf{a + \dfrac{1}{a} = 4 }$

☯TO FIND :

• $\bf{ a^4 + \dfrac{1}{ a^4} =? }$

☯FORMULA REQUIRED :

• $\bigstar \underline{ \boxed{ \red{\bf{{(a +b)}^{2} = a^2 + 2ab + b^2 }}}}$

➲SOLUTION :

• Squaring on both sides :

$\implies { \sf{{ \bigg(a + \dfrac{1}{a} \bigg)}^{2} = 4 ^2 }}$

$\implies { \sf{ {a}^{2} + 2 \times \cancel{a} \times \dfrac{1}{ \cancel{a}} + { \bigg( \dfrac{1}{a} \bigg )}^{2} = 16 }}$

$\implies { \sf{ {a}^{2} + 2 + \dfrac{1}{ a^2 } = 16 }}$

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

$\implies \underline{\boxed{ \bf{ {a}^{2} + \dfrac{1}{ a^2 } = 14}}}$

• Squaring on both sides (Again) :

$\implies { \sf{{ \bigg(a^2 + \dfrac{1}{a^2} \bigg)}^{2} = 14 ^2 }}$

$\implies { \sf{ {(a^2)}^{2} + 2 \times \cancel{a^2} \times \dfrac{1}{ \cancel{a^2}} + { \bigg( \dfrac{1}{a^2} \bigg )}^{2} = 196 }}$

$\implies { \sf{ {a}^{4} + 2 + \dfrac{1}{ a^4 } = 196 }}$

$\implies { \sf{ {a}^{4} + \dfrac{1}{ a^4 } = 196 - 2 }}$

$\implies \underline{\boxed{ \pink {\bf{ {a}^{4} + \dfrac{1}{ a^4} = 194}}}}$

$\huge{\green{\therefore}}$ $\bf{ a^4 + \dfrac{1}{ a^4} =194 }$