Fortunately, one of the three roots of 27y3 - 1 is really obvious, y = 1/3. Therefore we can pull (y - 1/3) out of the expression giving us:
(y - 1/3)(27y2 + 9y +3).
Using the quadratic formula on the second multiplicative expression gives:
[-9 ± √(-243)]/54 = (-9 ± 9i√3)/54.
So the complete, factored form of 27y3 - 1 is:
(y - 1/3)[y - (9 + 9i√3)/54][y - (9 - 9i√3)/54].
(x3 - 8) is factored thus: (x - 2)(x2 + 2x + 4) The easiest way to do this is to remember the formula: (a3 - b3) = (a - b)(a2 + ab + b2)
A cubed
5x5x5
Oh, dude, like, when you see p cubed in an expression, it means you're multiplying p by itself three times. So, it's like p * p * p. It's just a fancy math way of saying "Hey, let's make p hang out with itself a couple more times."
(x - 4)(x^2 + 4x + 16)
The number 6 can be factored into 2 x 3. So 6 cubed can be factored into 2 cubed times 3 cubed. In either case, the result is equal to 216.
-1
no
(t+3)3 is already factored. Could write it as (t+3)(t+3)(t+3).
It is 7*7*7.
The expression a^3 + b^3 can be factored using the sum of cubes formula, which states that a^3 + b^3 = (a + b)(a^2 - ab + b^2). Therefore, a^3 + b^3 can be factored as (a + b)(a^2 - ab + b^2). This formula helps us break down the sum of two cubes into a product of binomials, simplifying the expression.
5x5x5 5x5x5 5x5x5 5x5x5
2 x 2 x 2 = 8
(x3 - 8) is factored thus: (x - 2)(x2 + 2x + 4) The easiest way to do this is to remember the formula: (a3 - b3) = (a - b)(a2 + ab + b2)
A cubed
The expression is 42/23
2a^3 + 54 can first be factored into 2 (a^3 + 27). This can then be factored out to 2 (a + 3) (a^2 - 3a + 9).