If x2 + 1/x2 =7, find the value of X3 + 1/x3
Answers
Answer:
Answer
Ifx
2
+1/x
2
−7 find the valve of x
3
+1/3
3
Given equation is x
2
+1/x
2
=7
we know that(x+1/x)
2
=x
2
+2.x.1/x+(1/x)
2
=x
2
+2+1/x
2
=(x
2
+1/x
2
)+2
=7+2
=9.
∴(x+1/x)
2
=9
⇒(x+1/x)=
(3)
x
⇒(x+1/x)=3.
Now ,
(x
3
+1/x)
3
=(x+1/x)
3
−3.x.1/x.(x+1/x)
=(x+1/x)
3
−3(x+1/x)
=(3)
3
−3(3)
=27−9
=18.
∴x
3
+1/x
3
=18.
Answer:
Adding 2 on both sides
x^2 + 1/x^2 + 2 = 9
x^2 + 1/x^2 + 2*x*1/x = 9
(x + 1/x)^2 = 9
x + 1/x = 3 or x + 1/x = -3
Cubbing both sides on positive
x^3 + 1/x^3 + 3*x*1/x *(x + 1/x) = 27
x^3 + 1/x^3 + 3*(3) = 27
x^3 + 1/x^3 = 18 —(1)
Cubbing both sides on negative
x^3 + 1/x^3 + 3*x*1/x*(x+ 1/x) = -27
x^3 + 1/x^3 - 9 = -27
x^3 + 1/x^3 = -18—(2)
Given that
x^2+1/x^2 = 7
(x+1/x)^2 =x^2+1/x^2+2
= 7+2 =9
Therefore
(x+1/x)=3 or -3
Now,
(x+1/x)^3 = x^3+1/x^3 +3(x+1/x)
Replacing x+1/x with 3,
(3)^3 = x^3+1/x^3 +3(3)
27=x^3+1/x^3 +9
Therefore,
x^3+1/x^3 =27-9 =18
If x+1/x is replaced by -3,
x^3+1/x^3 would be -18.
x^2+1/x^2 = 7, or
x^4–7x^2+1 = 0 …(1)
Let x^2 = y, so (1) becomes
y^2–7y+1 = 0
y1 = [7+(49-4)^0.5]/2
= [7+45^0.5]/2 = 6.854101966 and x1 = 2.618033989 and x2 = -2.618033989
x1^3+1/x1^3 = 2.618033989^3+ 1/2.618033989^3 = 17.94427101 +1/17.94427101 = 18
x2^3+1/x2^3 = -2.618033989^3- 1/2.618033989^3 = -17.94427101 -1/17.94427101 = -18
Answer is +18 or - 18.
OR ELSE IN ANOTHER METHOD YOU CAN SEE
We know that
(a + b)² = a² + b² + 2ab
Therefore
(x + 1/x)² = x² + 1/x² + 2×x×1/x
or, (x + 1/x)² = 7 + 2 . . . . . . . . . . . . . . { x² + 1/x² = 7 is given}
or, (x + 1/x)² = 9
or, x + 1/x = √9
So, x + 1/x = ±3
We also know that
a³ + b³ + 3ab(a + b) = (a + b)³
Therefore
x³ + 1/x³ + 3 × x × 1/x (x + 1/x) = (x + 1/x)³
or, x³ + 1/x³ + 3(x+ 1/x) = (x + 1/x)³
or, x³ + 1/x³ = (x + 1/x)³ - 3(x + 1/x)
or, x³ + 1/x³ = (x + 1/x){ (x + 1/x)² - 3 }
or, x³ + 1/x³ = ±3{(±3)² - 3} . . . . . . . . . {Because x + 1/x = ±3}
or, x³ + 1/x³ = ±3{9 - 3}
or, x³ + 1/x³ = ±3 × 6
So, x³ + 1/x³ = ±18