Which long division problem can be used to prove the formula for factoring the difference of two perfect cubes? '+ ab +6²
a+b) a² - ab +6²
a+b a² +0a²b+Oab² – 83
a-ba? +Da+b+Dab2 – 67

Step-by-step explanation:

let the two perfect cubes be a³ and b³. Factring the  difference of these two perfect cubes we have;

a³ - b³

First we need to factorize (a-b)³

(a-b)³ = (a-b) (a-b)²

(a-b)³ = (a-b)(a²-2ab+b²)

(a-b)³ = a³-2a²b+ab²-a²b+2ab²-b³

(a-b)³ = a³-b³-2a²b-a²b+ab²+2ab²

(a-b)³ = a³-b³ - 3a²b+3ab²

(a-b)³  = (a³-b³) -3ab(a-b)

Then we will make a³-b³ the subject of the formula from the resultinh equation;

a³-b³ =  (a-b)³+ 3ab(a-b)

a³-b³  = a-b{(a-b)²+3ab}

a³-b³  = a-b{a²+b²-2ab+3ab}

a³-b³  = (a-b)(a²+b²+ab)

a³-b³  = (a-b)(a²+ab+b²)

The long division problem that can be used is (a-b)(a²+ab+b²)

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.

at least two

step-by-step explanation: