IF x2+x+1=0 then find (x3+1/x3)3
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Answered by
31
x² + x + 1 = 0
⇒ x² + x + 1 -x = 0 - x
⇒x² + 1 = -x
⇒(x² + 1)/x = -x/x
⇒ x + 1/x = -1
Using a³+b³ = (a+b)³ - 3ab(a+b)
x³ + 1/x³ = (x+1/x)³ - 3×x×1/x×(x+1/x)
Thus (x³ + 1/x³)³ = [(x+1/x)³ - 3×x×1/x×(x+1/x)]³
=[ (-1)³ - 3×(x×1/x)×(-1) ]³
=[ (-1)³ - 3×(1)×(-1) ]³
=[ -1 -3×(-1) ]³
=[ -1 + 3 ]³
=[2]³
=8
⇒ x² + x + 1 -x = 0 - x
⇒x² + 1 = -x
⇒(x² + 1)/x = -x/x
⇒ x + 1/x = -1
Using a³+b³ = (a+b)³ - 3ab(a+b)
x³ + 1/x³ = (x+1/x)³ - 3×x×1/x×(x+1/x)
Thus (x³ + 1/x³)³ = [(x+1/x)³ - 3×x×1/x×(x+1/x)]³
=[ (-1)³ - 3×(x×1/x)×(-1) ]³
=[ (-1)³ - 3×(1)×(-1) ]³
=[ -1 -3×(-1) ]³
=[ -1 + 3 ]³
=[2]³
=8
prabhushettisan:
how u got (x2+1)/x=-x/x
subtract x from both sides of (x² + x + 1 = 0)
and then divide both sides by x
why sir
It's the way to solve...you need to manipulate it into such a form which you can use later to get the answer..
by expanding x³ + 1/x³, i will get x + 1/x, so i converted it into that form
sir prove that "three times of any side of an equilateral triangle is equal to four times the squares of the altitude"
apply pythogorus theorem - u will get it. did u create a question?
s canu u please do it sir please
sir
Answered by
14
x² + x + 1 = 0
multiply with (x-1) both sides:
(x-1)(x² + x + 1) = (x-1) 0
x³ - 1 = 0
x³ = 1
x³ + 1/x³ = 1 + 1/1 = 2
Answer is 2³ = 8
multiply with (x-1) both sides:
(x-1)(x² + x + 1) = (x-1) 0
x³ - 1 = 0
x³ = 1
x³ + 1/x³ = 1 + 1/1 = 2
Answer is 2³ = 8
sir prove that "three times of any side of an equilateral triangle is equal to four times the squares of the altitude"
sir why did u multiply (x-1) on both sides and u rote x3-1 on rhs
see now.
did u create a question for equilateral triangle? u can prove it by Pythagoras theorem.
let the altitude from A meet BC at D. AD is perpendicular to DC. in ADC, AD² + DC² = AC². AD² = AC² - (AC/2)² = 3/4 AC².
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