Advertisements
Advertisements
Question
If x = `1/[ 5 - x ] "and x ≠ 5 find "x^3 + 1/x^3`
Advertisements
Solution
Given x = `1/[ 5 - x ]`;
By cross multiplication
⇒ x(5 - x) = 1
⇒ x2 - 5x = -1
⇒ x2 + 1 = 5x
⇒ `[ x^2 + 1]/x = 5`
⇒ `[ x + 1/x ] = 5` ...(1)
We know that
`( x^3 + 1/x^3 ) = ( x + 1/x )^3 - 3( x + 1/x )`
= `(5)^3 - 3(5)` ...[From equation (1)]
= `x^3 + 1/x^3`
= 125 - 15
= 110
APPEARS IN
RELATED QUESTIONS
Expand : ( x + 8 ) ( x + 10 )
Expand : ( X - 8 ) ( X + 10 )
Expand: `( 2x - 1/x )( 3x + 2/x )`
Expand : ( x + y - z )2
Expand : ( 5x - 3y - 2 )2
If a + b + c = p and ab + bc + ca = q ; find a2 + b2 + c2.
If a + `1/a` = m and a ≠ 0; find in terms of 'm'; the value of `a^2 - 1/a^2`.
If 2( x2 + 1 ) = 5x, find :
(i) `x - 1/x`
(ii) `x^3 - 1/x^3`
If a2 + b2 = 34 and ab = 12; find : 7(a - b)2 - 2(a + b)2
If x2 + `x^(1/2)`= 7 and x ≠ 0; find the value of:
7x3 + 8x − `7/x^3 - 8/x`
