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the answer is (3x-2)(9x squared+6x+4)

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What is 3x cubed times 3x cubed?

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


How do you factor 20x squared minus 27x minus 8?

20x2 - 27x -8 20 * -8 is -160, and -32 and 5 add to -27 20x2 - 32x + 5x - 8 4x(5x-8) + 1(5x-8) (4x + 1)(5x-8)


2x3 minus 3x2 minus 11x plus 7 divided by x plus 3?

2x^3-3x^2-11x+7x+32x^2-9x+16 R -41x+3 l 2x^3-3x^2-11x+7-(2x^3+6x^2)-9x^2-11x+7-(-9x^2-27x)16x+7-(16x+48)-412x^2-9x+16-[41/(x+3)]


Suppose that y varies directly as the cube of x If x is tripled what is the effect on y?

y will be multiplied by 27. Remember, whatever you do to one side of an equation, you must do to the other. In this case, the relationship between x and y would be expressed as y = x^3 To triple x would look like y = (3x)^3 ...b/c you're tripling the value x alone, not the entire term x^3 y = 27x^3 ...after distributing the exponent (3^3 = 27) In order to set both sides of the equation equal, y must be multiplied by 27... 27y = 27x^3 TADA!


A closed rectangular box is to be constructed with a base as long as its wide suppose that the total surface area is 27square feet find the dimension of the box that maximize the volume?

Since the base is equal to the length, then the two parts of the box, up and down are two squares, let's say with length side x. So the other four parts may be rectangles with length x, and wide y, (where y is also the height of the box). Since the surface area is 27 ft^2, we have: Surface Area = 27 = 2x^2 + 4xy Solve for y y = (27 - 2x^2)/4xThus, we have:l = xw = xh = (27 - 2x^2)/4xV = lwhV= (x)(x)[27 - 2x^2)/4x] Simplify and multiply:V = (27x - 2x^3)/4V = (1/4)(27x - 2x^3) Take the derivative: V'= (1/4)(27 - 6x^2) Find the critical values by setting the derivative equal to zero:0 = (1/4)(27 - 6x^2) multiply by 0 both sides:0 = 27 - 6x^2 add 6x^2 to both sides:6x^2 = 27 divide by 6 to both sides:x^2 = 27/6x^2 = 9/2 Square both sides:x = √(9/2)x = (3√2)/2y = (27 - 2x^2)/4xy = 27/4x - (2x^2)/4xy = 27/4x - x/2 substitute (3√2)/2 for x:y = 27/][4(3√2)/2)] - (3√2)/2)/2y = (9√2)/4 - (3√2)/4y = (6√2)/4y = (3√2)/2 As we see the box is a cube.V = side^3V = [(3√2)/2]^3 V = (27√2)/4So, V = (27√2)/4 ft^3 when x = (3√2)/2. Thus, we can maximize the volume of this box if l = w = h = (3√2)/2 ft.

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