If [x^2 +(1/x^2)=7] what is the value of [7x^3+8x-(7/x^3)-(8/x)]I know the answer which is ±64 But I want a step-by-step explaination on this question.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find 7x^3 + 8x - (7/x^3) - (8/x) given that x^2 + 1/x^2 = 7

7x^3 + 8x - (7/x^3) - (8/x)

=> 7(x^3 - 1/x^3) + 8(x - 1/x)

=> 7(x - 1/x)(x^2 + x*(1/x) + 1/x^2) + 8(x - 1/x)

=> (x - 1/x)[7*(x^2 + x*(1/x) + 1/x^2) + 8]

=> (x - 1/x)[7x^2 + 7 + 7/x^2 + 8]

=> (x - 1/x)[7(x^2 + 1/x^2) + 15]

as x^2 + 1/x^2 = 7

=> (x - 1/x)(7*7 + 15]

=> (x - 1/x)*64

Now, we have to determine x - 1/x

(x - 1/x)^2 = x^2 - 2*x*(1/x) + 1/x^2

=> x^2 - 2 + 1/x^2

=> 7 - 2

=> 5

x - 1/x = +sqrt 5 and x - 1/x = -sqrt 5

(x - 1/x)*64 = 64*sqrt 5 and -64*sqrt 5

The required value of 7x^3 + 8x - (7/x^3) - (8/x) is 64*sqrt 5 and -64*sqrt 5.

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll factorize by 7 the terms 7x^3 and 7/x^3:

7(x^3 - 1/x^3)

We'll factorize by 8 the terms 8x and 8/x:

8(x - 1/x)

We'll apply the identity:

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

(x^3 - 1/x^3) = (x - 1/x)(x^2 + x/x + 1/x^2)

But, form enunciation, x^2 + 1/x^2 = 7.

(x^3 - 1/x^3) = (x - 1/x)(7 + 1)

(x^3 - 1/x^3) = 8(x - 1/x)

7*(x^3 - 1/x^3) = 7*8(x - 1/x)

The expression to be calculated will become:

7(x^3 - 1/x^3) + 8(x - 1/x) = 7*8(x - 1/x) + 8(x - 1/x)

We'll factorize by 8(x - 1/x):

7(x^3 - 1/x^3) + 8(x - 1/x) = 8(x - 1/x)*(7+1)

7(x^3 - 1/x^3) + 8(x - 1/x) = 64(x - 1/x)

We'll calculate x - 1/x:

We'll raise to square x - 1/x:

(x - 1/x)^2 = x^2 - 2x/x + 1/x^2

But x^2 + 1/x^2 = 7.

(x - 1/x)^2 = 7 - 2

(x - 1/x)^2 = 5 => x - 1/x = sqrt 5

The value of expression will be:

7(x^3 - 1/x^3) + 8(x - 1/x) = 64*sqrt 5

The value of the given expression is:7x^3+8x-(7/x^3)-(8/x) = 64*sqrt 5.

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