if x square + 1 upon x square is equal to 62 find x cube + 1 upon x cube
Answers
Answer:
x
3
+
x
3
1
=488
Step-by-step explanation:
We have,
x^{2}+\dfrac{1}{x^2}=62x
2
+
x
2
1
=62
To find, the value ofx^{3}+\dfrac{1}{x^3}=?x
3
+
x
3
1
=?
∴ (x+\dfrac{1}{x})^{2} =x^{2}+\dfrac{1}{x^2}+2(x+
x
1
)
2
=x
2
+
x
2
1
+2
⇒ (x+\dfrac{1}{x})^{2}=64(x+
x
1
)
2
=64
⇒ (x+\dfrac{1}{x})^{2}=8^{2}(x+
x
1
)
2
=8
2
⇒ x+\dfrac{1}{x}=8x+
x
1
=8
∴ (x+\dfrac{1}{x})^{3} =x^{3}+\dfrac{1}{x^3}+3.x.\dfrac{1}{x}(x+\dfrac{1}{x})(x+
x
1
)
3
=x
3
+
x
3
1
+3.x.
x
1
(x+
x
1
)
⇒ (x+\dfrac{1}{x})^{3} =x^{3}+\dfrac{1}{x^3}+3(x+\dfrac{1}{x})(x+
x
1
)
3
=x
3
+
x
3
1
+3(x+
x
1
)
⇒ x^{3}+\dfrac{1}{x^3}=(x+\dfrac{1}{x})^{3}-3(x+\dfrac{1}{x})x
3
+
x
3
1
=(x+
x
1
)
3
−3(x+
x
1
)
⇒ x^{3}+\dfrac{1}{x^3}=(8)^{3}-3(8)=512-24x
3
+
x
3
1
=(8)
3
−3(8)=512−24
⇒x^{3}+\dfrac{1}{x^3}=488x
3
+
x
3
1
=488
Hence, x^{3}+\dfrac{1}{x^3}=488x
3
+
x
3
1
=488
Answer:
232 or -280
Step-by-step explanation:
as (x+1/x)^2= x^2+ 1/x^2 + 2
putting 62 we get x+1/x = +8 and -8
we need to consider both values unless mentioned
now put the valu in
(x+1/x)^3 = x^3+ 1/x^3 +3(x+1/x)
for x+1/x =8
ans is 232
and for
x+1/x = -8
ans is -280