cube root of 27 = 3, so 3^12 = 531 441 * * * * * If the question is about 27x, rather than 27, the answer is as follows: cuberoot(27x12) = (27x)12/3 = (27x)4 = 531441x4
27X^3-8 = 0 27X^3 = 8 X^3 = 8/27 X = cubic root of 8/cubic root of 27 = 2/3
It is: 23-27x when simplified
(8x-8y)/(8x+8y) = (x-y)/(x+y)Cancel down by 8; you have simplified the equation.
If you mean (9x-8y)(9x+8y) then it is 81x^2 minus 64y^2
The LCM is 72y^3.
72y
cube root of 27 = 3, so 3^12 = 531 441 * * * * * If the question is about 27x, rather than 27, the answer is as follows: cuberoot(27x12) = (27x)12/3 = (27x)4 = 531441x4
Your question is ambiguous. (3x)^3 times (3x)^3 = 27x^3 times 27x^3 = 729x^6 or, if you meant 3x^3 times 3x^3 = 9x^6
The LCM is 40y3
Not sure which you mean, take your pick: (8y)3+27=(8y+3)((8y)2-3*(8y)+9)=(8y+3)(64y2-24y+9); or 8y3+27=(2y+3)(4y2-6y+9)
27X^3-8 = 0 27X^3 = 8 X^3 = 8/27 X = cubic root of 8/cubic root of 27 = 2/3
the answer is (3x-2)(9x squared+6x+4)
(8x + 9)(3x^2 - 1)
I also need to know the answer to this problem. Can anyone answer it?
8(y - 1)(y^2 + y + 1)
In order to factor the sum of the cubes, we need to use this form a³ + b³ = (a + b)(a² - ab + b²). Let a³ = 27x³ and b³ = 343y³. Then, a = 3x and b = 7y. Perform substitution of these terms for the form, and we obtain: 27x3 + 343y3 = (3x + 7y)(9x2 - 21xy + 49y2)