Question

x+1/x=7
Then what is x7+1/x7.
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Answer 1

Answer:

x + 1 / x = 7

We know the algebraic identity,

a³ + b³ + 3ab ( a + b ) = ( a + b )³

Or

a³ + b³ = ( a + b )³ - 3ab( a + b )

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x³ + 1/x³ = (x + 1/x )³- 3 × x ×1/x(x + 1/x )

= ( x + 1/x )³ - 3( x + 1/x )

= 7³ - 3 × 7

= 7( 7² - 3 )

= 7 ( 49 - 3 )

= 7 × 46

= 322

We will be using the Identity of ( a + b) ³

Given,

x + 1/x = 7 .

Now,

Cubing on both sides,

( x + 1/x)³ = 7³

x³ + 1/x³ + 3x(1/x) ( x + 1/x) = 343

x³ + 1/x³ = 343 - 3 ( 7)

x³ + 1/x³ = 322

Assuming the question is “If x-(1/x)=4, then what is x^7-(1/x)^7=?” as otherwise it would be a very easy problem.

We can factorize a difference of 7th powers using the identity (easily verified by expanding the brackets):

a7−b7=(a−b)(a6+a5b+a4b2+a3b3+a2b4+ab5+b6)

Similarly, taking x=a and 1/x=b, we can write the desired expression:

x7−1x7=(x−1x)(x6+x4+x2+1+1x2+1x4+1x6) ……(1)

We then need to calculate the values of three different partial sums:

x2+1x2, x4+1x4, and x6+1x6

Squaring the given equation, x-(1/x)=4, we have:

(x−1x)2=16=x2−2+1x2⟹

x2+1x2=18 …………………..……….(2)

⟹x4+1x4=322

Multiplying above expressions (2) and (3) yields the sum of 6th powers:

(x2+1x2)(x4+1x4)=18×322

⟹x6+1x6+x2+1x2=18×322

⟹x6+1x6=5778

Replacing (2), (3), and (4) into (1), we have:

x7−1x7=4(5778+322+18+1)=24476