Need a little math help regarding exponents and stuff.?

x= 2^(1/3) + 2^(2/3)

VALUE OF

f(x) = x^3 - 64

[2^(1/3) + 2^(2/3)]^3 - 64

[2^(1/3) + 2^(2/3)][2^(1/3) + 2^(2/3)][2^(1/3) + 2^(2/3)] - 64

[2^(1/3) + 2^(2/3)][(2^(1/3+2/3)) + 2^(1/3+2/3) +2^(1/3+1/3) + 2^(2/3+2/3)] - 64

[2^(1/3) + 2^(2/3)][2^1 + 2^1 + 2^(2/3) + 2^(4/3)] - 64

[2^(1/3) + 2^(2/3][4 + 2^(2/3) + 2^(4/3)] - 64

4(2^(1/3)) + 2^(1/3+2/3) + 2^(1/3+4/3) + 4(2^(2/3)) + 2^(2/3+2/3) + 2^(2/3+4/3) - 64

4(2^(1/3)) + 2^1 + 2^5/3 + 4(2^(2/3)) + 2^(4/3) + 2^(6/3) - 64

4(2^(1/3)) + 2 + 2^(5/3) + 4(2^(2/3)) + 2^(4/3) + 2^2 - 64

4(2^(1/3)) + 2^(5/3) + 4(2^(2/3)) + 2^(4/3) + 2 +4 - 64

4(2^(1/3)) + 4(2^(2/3)) + 2^(5/3) + 2^(4/3) - 58

4(^3√(2)^1) + 4(^3√(2)^2) + ^3√(2)^5 + ^3√(2)^4 - 58

4(^3√2) + 4(^3√4) + ^3√32 + ^3√16 - 58

OPTION D

interior the 1st component of the equation, you may sq. the two factors by using x, which will cancel out the x. this could pass away you with: z = y^2 Now that that's ordinary, initiate plugging in numbers with suitable squares. F(4): 4 = 2^2 F(4): 2^4 = 2 * 4^x F(4): sixteen = 2 * 4^x F(4): 8 = 4^x F(4): x = a million.5 a million.5 + 2 + 4 = 7.5 z =/= 4 ------------------------------- F(9): 9 = 3^2 F(9): 2^9 = 2 * 4^x F(9): 512 = 2 * 4^x F(9): 256 = 4^x F(9): x = 4 4 + 3 + 9 = sixteen ^ answer ^

5^x-3 * 3^2x-8=225

Given the numbers available, it could be just trial and error. It so happens to be 5.

[5^(5-3) * 3^(2(5)-8)=225]

5^2 * 3^2= 225

Enjoy :)