if x² +1/25x²=43/5 find the value of x³+1/125x³
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x² + 1/25x² = 43/5
x³ + 1/125x³ = ?
( x + 1/5x )² = x² + 1/25x² + 2/5
( x + 1/5x )² = 43/5 + 2/5
( x + 1/5x )² = (43 + 2)/5
( x + 1/5x )² = 45/5
( x + 1/5x )² = 9
x + 1/5x = √9
x + 1/5x = 3 or - 3[Since, 3² = 9 and (-3)² = 9 too]
When, x + 1/5x = 3,
( x + 1/5x )³ = ( 3 )³
x³ + 1/125x³ + 3/5( x + 1/5x ) = 27
x³ + 1/125x³ + (3/5) * 3 = 27
x³ + 1/125x³ = 27 - 9/5
x³ + 1/125x³ = ( 135 - 9 )/5
x³ + 1/125x³ = 126/5 = 25.2
When, x + 1/5x = - 3,
( x + 1/5x )³ = ( - 3 )³
x³ + 1/125x³ + 3/5( x + 1/5x ) = - 27
x³ + 1/125x³ + 3/5( - 3 ) = - 27
x³ + 1/125x³ = - 27 + 9/5
x³ + 1/125x³ = - 126/5 = - 25.2
Therefore,
x³ + 1/125x³ = 25.2 or - 25.2
yash141799:
x³ + 1/125x³ + 3/5( x + 1/5x ) = 27 why is 3/5 placed here?
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