Explanation:
Given: x = {√(2) + 1}^(-1/3)
Asked: x³+(1/x³) = ?
Solution:
We have, x = {√(2) + 1}^(-1/3)
⇛x = [1/{√(2) + 1}^(1/3)]
[since, (a⁻ⁿ = 1/aⁿ)]
Cubing on both sides, then
⇛(x)³ = [1{/√(2) + 1}^(1/3)]³
⇛(x)³ = [(1)³/{√(2) + 1}^(1/3 *3)]
⇛(x)³ = [(1)³/{√(2) + 1}^(1*3/3)]
⇛(x)³ = [(1)³/{√(2) + 1}^(3/3)]
⇛(x * x * x) = [(1*1*1)/{√(2) + 1)^1]
⇛x³ = [1/{√(2) + 1}]
Here, we see that on RHS, the denominator is √(2)+1. We know that the rationalising factor of √(a)+b = √(a)-b. Therefore, the rationalising factor of √(2)+1 = √(2) - 1. On rationalising the denominator them
⇛x³ = [1/{√(2) + 1}] * [{√(2) - 1}/{√(2) - 1}]
⇛x³ = [1{√(2) + 1}/{√(2) + 1}{√(2) - 1}]
Multiply the numerator with number outside of the bracket with numbers on the bracket.
⇛x³ = [{√(2) + 1}/{√(2) + 1}{√(2) - 1}]
Now, Comparing the denominator on RHS with (a+b)(a-b), we get
Using identity (a+b)(a-b) = a² - b², we get
⇛x³ = [{√(2) - 1}/{√(2)² - (1)²}]
⇛x³ = [{√(2) - 1}/{√(2*2) - (1*1)}]
⇛x³ = [{√(2) - 1}/(2-1)]
⇛x³ = [{√(2) - 1}/1]
Therefore, x³ = √(2) - 1 → → →Eqn(1)
Now, 1/x³ = [1/{√(2) - 1]
⇛1/x³ = [1/{√(2) - 1] * [{√(2) + 1}/{√(2) + 1}]
⇛1/x³ = [1{√(2) + 1}/{√(2) - 1}{√(2) + 1}]
⇛1/x³ = {√(2) + 1}/[{√(2) - 1}{√(2) + 1}]
⇛1/x³ = [{√(2) + 1}/{√(2)² - (1)²}]
⇛1/x³ = [{√(2) + 1}/{√(2*2) - (1*1)}]
⇛1/x³ = [{√(2) + 1}/(2-1)]
⇛1/x³ = [{√(2) + 1}/1]
Therefore, 1/x³ = √(2) + 1 → → →Eqn(2)
On adding equation (1) and equation (2), we get
x³ + (1/x³) = √(2) -1 + √(2) + 1
Cancel out -1 and 1 on RHS.
⇛x³ + (1/x³) = √(2) + √(2)
⇛x³ + (1/x³) = 2
Therefore, x³ + (1/x³) = 2
Answer: Hence, the required value of x³ + (1/x³) is 2.
Please let me know if you have any other questions.