%5E2%3D3-show-that-(x%5E3%2B1%2Fx%5E3)%3D0.jpeg "(x+1/x)^2=3 Show That (x^3+1/x^3)=0")
Proving (x³ + 1/x³) = 0 given (x + 1/x)² = 3
This problem involves manipulating algebraic expressions and utilizing the given equation to arrive at the desired result. Here's how to prove it:
1. Expand the given equation:
(x + 1/x)² = 3 x² + 2(x)(1/x) + (1/x)² = 3 x² + 2 + 1/x² = 3
2. Isolate the term (x² + 1/x²):
x² + 1/x² = 3 - 2 x² + 1/x² = 1
3. Cube the equation from step 2:
(x² + 1/x²)³ = 1³ x⁶ + 3(x²)(1/x²) + 3(x²)(1/x⁴) + 1/x⁶ = 1 x⁶ + 3 + 3(1/x²) + 1/x⁶ = 1
4. Simplify the equation:
x⁶ + 1/x⁶ + 3 + 3/x² = 1 x⁶ + 1/x⁶ + 3(1 + 1/x²) = 1
5. Substitute the value of (x² + 1/x²) from step 2:
x⁶ + 1/x⁶ + 3(1) = 1 x⁶ + 1/x⁶ + 3 = 1
6. Isolate the term (x⁶ + 1/x⁶):
x⁶ + 1/x⁶ = 1 - 3 x⁶ + 1/x⁶ = -2
7. Factor the term (x⁶ + 1/x⁶):
(x³ + 1/x³)(x³ - 1/x³) = -2
8. Observe that the term (x³ - 1/x³) is non-zero.
This is because if it were zero, then (x⁶ + 1/x⁶) would also be zero, which contradicts our result in step 6.
9. Therefore, (x³ + 1/x³) must be equal to zero.
This is the only way for the product of two terms to be -2, while one term is non-zero.
Therefore, we have successfully shown that (x³ + 1/x³) = 0 given (x + 1/x)² = 3.