(please help me to solve this question)
solving equation
(1+x)^2/3 + (1-x)^2/3 = 4(1-x²)^1/3 are,
value of x is
To solve the equation (1+x)^2/3 + (1-x)^2/3 = 4(1-x^2)^1/3, we need to isolate the variable x.
Let's start by simplifying the equation.
First, let's use the property of exponents that states (a^m)^n = a^(m*n).
So, (1+x)^2/3 = [(1+x)^2]^(1/3) and (1-x)^2/3 = [(1-x)^2]^(1/3).
Applying this property, the equation becomes:
[(1+x)^2]^(1/3) + [(1-x)^2]^(1/3) = 4(1-x^2)^(1/3).
Next, we can let a = (1+x)^2 and b = (1-x)^2, which simplifies the equation further:
a^(1/3) + b^(1/3) = 4(1 - x^2)^(1/3).
Now, raising both sides of the equation to the power of 3, we get:
[a^(1/3) + b^(1/3)]^3 = [4(1 - x^2)^(1/3)]^3.
Expanding the left side using the binomial theorem, we have:
a + 3a^(2/3)b^(1/3) + 3ab^(2/3) + b = 64(1 - x^2).
Now, substitute a = (1+x)^2 and b = (1-x)^2 back into the equation:
(1+x)^2 + 3(1+x)^2/3(1-x)^2/3(1-x)^2 + 3(1+x)^2(1-x)^2/3 + (1-x)^2 = 64(1 - x^2).
This equation may look complex, but we can simplify it further.
Let's simplify the terms: (1+x)^2 and (1-x)^2.
(1+x)^2 = 1 + 2x + x^2, and (1-x)^2 = 1 - 2x + x^2.
Now, substitute these expressions back into the equation:
1 + 2x + x^2 + 3[1 + 2x+x^2]^(2/3)[1-2x+x^2]^(1/3) + 3[1 + 2x + x^2][1 - 2x + x^2]^(2/3) + 1 - 2x + x^2 = 64(1 - x^2).
Simplify the exponents in the middle terms:
1 + 2x + x^2 + 3[1 + 2x + x^2]^(2/3)[1 - 2x + x^2]^(1/3) + 3[1 + 2x + x^2]^(1/3)[1 - 2x + x^2]^(2/3) + 1 - 2x + x^2 = 64(1 - x^2).
Now, combine like terms and simplify further.
After simplifying, we are left with:
4x + 6[x^2 (1 - x^2)]^(1/3) = 0.
This equation implies that either 4x = 0 or 6[x^2 (1 - x^2)]^(1/3) = 0.
For 4x = 0, x must be equal to 0.
For 6[x^2 (1 - x^2)]^(1/3) = 0, we need to solve for x^2 = 0 or (1-x^2) = 0.
If x^2 = 0, x can be either 0 or -0.
If (1-x^2) = 0, solving for x, we have x = 1 or x = -1.
Therefore, the possible values of x that satisfy the equation are x = -1, 0, and 1.