Question

if x=(4+√15)^1/3+(4-√15)^1/3 then show that x^3-3x-8=0​

Answer 1

Step-by-step explanation:

here is ur answer

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Answer 2

x^3-3x-8=0

Given:

x=(4+√15)^1/3+(4-√15)^1/3

To Show:

x^3-3x-8=0

Solution:​

$x = (4+\sqrt{15} )^{\frac{1}{3} } + (4-\sqrt{15} )^{\frac{1}{3} }$

Cubing on both sides

$x^{3} = [(4+\sqrt{15} )^{\frac{1}{3} } + (4-\sqrt{15} )^{\frac{1}{3} }]^{3}$

∵ Formula (A+B)³ = A³ +B³+3AB(A+B)

Here A= $(4+\sqrt{15} )^{\frac{1}{3}$ and B = $(4-\sqrt{15} )^{\frac{1}{3}$

$x^{3} = [(4+\sqrt{15} )^{\frac{1}{3} }]^{3} + [(4-\sqrt{15} )^{\frac{1}{3} }]^{3} + 3*(4+\sqrt{15} )^{\frac{1}{3} } * (4-\sqrt{15} )^{\frac{1}{3} }*[ (4+\sqrt{15} )^{\frac{1}{3} } + (4-\sqrt{15} )^{\frac{1}{3} }]$

∵ A^m+ B^m = (A+B)^m

$x^{3} = 4+\sqrt{15} + 4-\sqrt{15} ) + 3*[(4+\sqrt{15} ) (4-\sqrt{15}} ]^{\frac{1}{3}} * [(4+\sqrt{15} )^{\frac{1}{3}}+ (4-\sqrt{15} )^{\frac{1}{3}}]$

∵ (A+B)(A-B) =A²-B²

$x^{3} = 8 + 3* (16-15)^{\frac{1}{3} } * x -------( Given)$

$x^{3}= 8+3*1*x\\\\x^{3}= 8 +3x\\\\x^{3}-8 -3x=0$

Thus Showed.

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