Question

please solve if you are (god) in factor​

Answer 1

hey !!

x+1/x = 2 ( cubing both sides )

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

(x+1/x ) ^3 = 8. ( According to the formula ,

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

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

x^3 + (1/x) ^3 + 3(x + 1/x ) = 8 ( as x+ 1/x = 2 )

x^3 + (1/x) ^3 + 3(2) = 8

x^3 + (1/x) ^3 + 6 = 8

x^3 + (1/x) ^3 = 8-6

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

hope it helps u...

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Answer 2

Answer:

let the value of x+1/x =2

therefore by cubing this x+1/x we get

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

=>(2)3 = x3 + 3 (2) + 1/x3 ( by substituting the value given above we get)

=> 6 = x3 + 1/x3 +6

=> 6-6 = x3 + 1/x3

=> 0 = x3 + 1/x3

therefore the value of x3 + 1/x3 = 0

तथा x3 + 1/x3 ,. 0 के बराबर है।

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