If x*3+1/x*3=110 find the value of x-1/x. The answer of this question is 5
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Heya..!!!
X³ + 1/x³ = 110
We know that:-
(a + b)³ = a³ + b³ + 3ab(a + b)
Here, a = x , b = 1/x
⇒(x + 1/x)³ = x³ + 1/x³ + 3*x *1/x (x + 1/x)
∴ [ let (x + 1/x) = p]
⇒(p)³ = 110 + 3p
∴ [ x³ + 1/x³ = 110 ]
⇒p³ - 3p - 100 = 0
⇒p³ + 0p² - (25 - 22)p - 110 = 0
∴[ 0 can be written as 5 - 5 = 0 ]
⇒p³ + 5p² - 5p² - 25p + 22p - 110 = 0
⇒ P³ + 5p² + 22p - 5p² - 25p - 110 = 0
⇒p(p² + 5p + 22) - 5(p² + 5p + 22) = 0
⇒(p² + 5p + 22) (p - 5) = 0
⇒(p - 5) = 0 or (p² + 5p + 22) = 0
⇒p = 5 or (p² + 5p + 22) = 0 ∴ [ neglect]
now p = 5
Hence,
X + 1/x = 5
⇒(x + 1/x)³ =( x³ + 1/x³) + 3*x *1/x (x + 1/x)
∴ [ x³ + 1/x³ = 110 ]
⇒ (5)³ = 110 + 3(5)
⇒125 = 110 + 15
⇒125 = 125
I HOPE ITS HELP YOU,
X³ + 1/x³ = 110
We know that:-
(a + b)³ = a³ + b³ + 3ab(a + b)
Here, a = x , b = 1/x
⇒(x + 1/x)³ = x³ + 1/x³ + 3*x *1/x (x + 1/x)
∴ [ let (x + 1/x) = p]
⇒(p)³ = 110 + 3p
∴ [ x³ + 1/x³ = 110 ]
⇒p³ - 3p - 100 = 0
⇒p³ + 0p² - (25 - 22)p - 110 = 0
∴[ 0 can be written as 5 - 5 = 0 ]
⇒p³ + 5p² - 5p² - 25p + 22p - 110 = 0
⇒ P³ + 5p² + 22p - 5p² - 25p - 110 = 0
⇒p(p² + 5p + 22) - 5(p² + 5p + 22) = 0
⇒(p² + 5p + 22) (p - 5) = 0
⇒(p - 5) = 0 or (p² + 5p + 22) = 0
⇒p = 5 or (p² + 5p + 22) = 0 ∴ [ neglect]
now p = 5
Hence,
X + 1/x = 5
⇒(x + 1/x)³ =( x³ + 1/x³) + 3*x *1/x (x + 1/x)
∴ [ x³ + 1/x³ = 110 ]
⇒ (5)³ = 110 + 3(5)
⇒125 = 110 + 15
⇒125 = 125
I HOPE ITS HELP YOU,
abhi569:
Sir ji,, we have to find the value of
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