Step-by-step explanation:
<u>Given</u>: {x+(1/x)}³ = 3
<u>Asked</u>: x³ + (1/x³) = ?
<u>Solution:</u>
<u>Method</u><u> </u><u>1:</u>
We have, {x+(1/x)}³ = 3
Comparing the expression with (a+b)³, we get
a = x
b = (1/x)
Using identity (a+b)³ = a³+b³+3ab(a+b), we get
⇛{x+(1/x)}³ = 3
⇛(x)³ + (1/x)³ + 3(x)(1/x){x + (1/x)} = 3
⇛(x*x*x) + (1*1*1/3*3*3) + 3(x)(1/x){x + (1/x)} = 3
⇛x³ + (1/x³) + 3(x)(1/x){x + (1/x)} = 3
⇛x³ + (1/x³) + 3{x + (1/x)} = 3
⇛x³ + (1/x³) + 3(x) + 3(1/x) = 3
⇛x³ + (1/x³) + 3x + (3/x) = 3
Our answer came incorrect.
Let's try..
<u>Method</u><u> </u><u>2</u><u>:</u>
We have,
[x+(1/x)]³ = 3
On taking cube root both sides then
⇛³√[{ x+(1/x)}³ ] = ³√3
⇛x+(1/x) = ³√3 -----(1)
We know that
a³+b³ = (a+b)³-3ab(a+b)
⇛x³+(1/x)³ = [x+(1/x)]³ - 3(x)(1/x)[x+(1/x)]
⇛x³+(1/x³) = (3)-3(1)(³√3)
[since, {x + (1/x)} = ³√3 from equation (1)]
⇛x³+(1/x)³ = 3-3 ׳√3
⇛x³ + (1/x³) = 3- ³√81 (or )
⇛x³ + (1/x³) = 3(1-³√3)
Therefore, x³ + (1/x³) = 3(1 - cube root of 3)
It is impossible to get zero
