Question

A^2+1/a^2=11 then the value of a^3-1/a^3=

Answer 1

Question:

If a^2 + 1/a^2 = 11 , then find the value of a^3 - 1/a^3 .

Answer:

a^3 - 1/a^3 = 36

Note:

• (A+B)^2 = A^2 + B^2 + 2•A•B

• (A-B)^2 = A^2 + B^2 - 2•A•B

• A^2 - B^2 = (A+B)(A-B)

• A^3 - B^3 = (A-B)(A^2 + B^2 + A•B)

• A^3 + B^3 = (A+B)(A^2 + B^2 - A•B)

Solution:

We have;

a^2 + 1/a^2 = 11 --------(1)

We know that;

(A-B)^2 = A^2 + B^2 - 2•A•B

Thus,

=> (a -1/a)^2 = a^2 + (1/a)^2 - 2•a•(1/a)

=> (a -1/a)^2 = a^2 + 1/a^2 - 2•a•(1/a)

=> (a -1/a)^2 = 11 - 2 {using eq-(1)}

=> (a -1/a)^2 = 9

=> (a -1/a) = √9

=> (a -1/a) = 3 -------------(2)

Also,

We know that,

A^3 - B^3 = (A-B)(A^2 + B^2 + A•B)

Thus;

=> a^3-1/a^3=(a-1/a){a^2+1/a^2+a(1/a)}

=> a^3 - 1/a^3 = 3•(11 + 1)

=> a^3 - 1/a^3 = 3•12

=> a^3 - 1/a^3 = 36

Hence,

The required value of a^3 - 1/a^3 is 36.

Answer 2

Answer:-

$\implies \: \boxed{ {a}^{3} - \frac{1}{ {a}^{3} } = 36} \:$

Step - by - step explanation:-

Used identities :-

$\star \: \red{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy} \\ \\ \star \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy)$

Solution :-

According to the question,

$\: \star \: {a}^{2} + \frac{1}{ {a}^{2} } = 11$

We know that,

${ \bigg(a - \frac{1}{a} \: \bigg) }^{2} = {a}^{2} + \frac{1}{ {a}^{2} } - 2 \times a \times \frac{1}{a} \\ \\ \because \: {a}^{2} + \frac{1}{ {a}^{2} } = 11 \\ \therefore \: { \bigg(a - \frac{1}{a} \: \bigg) }^{2} \: = 11 - 2 \\ \\ \implies \: { \bigg(a - \frac{1}{a} \: \bigg) }^{2} \: = 9 \\ \\ \bf{ taking \: \sqrt{} \: \: on \: both \: sides} \\ \\ \implies \: \sqrt{ { \bigg(a - \frac{1}{a} \bigg) }^{2} } = \sqrt{9} \\ \\ \boxed{ \implies \: a - \frac{1}{a} = 3}$

Now ,

${a}^{3} - \frac{1}{ {a}^{ 3} } = (a - \frac{1}{a} )( {a}^{2} + \frac{1}{ {a}^{2} } + a \times \frac{1}{a} ) \\ \\ \because \: a - \frac{1}{a} = 3 \: \: \: and \: {a}^{2} + \frac{1}{ {a}^{2} } = 11 \\ \\ \therefore \: {a}^{3} - \frac{1}{ {a}^{3} } = 3(11 +1) \\ \\ \implies \: {a}^{3} - \frac{1}{ {a}^{3} } = 3 \times 12\\ \\ \implies \: \boxed{ {a}^{3} - \frac{1}{ {a}^{3} } = 36}$