Question

The product of 3 consecutive even integers is equal to the cube of the first plus the square of the second plus twice the square of the third. Find the integers. Please show work.

Answer 1

6, 8 and 10 and -2, 0, 2

There are two roots x = -2 and 6,

but the since I don't believe 0 is an even number the answer is 6, 8 and 10

I was incorrect according to G0ggle zero is an even number so another answer is -2, 0, 2

Explanation:

read the question and convert the English to mathematics

x, y and z even consecutive number

y = x+2

z = y+2 = x+4

and

xyz = x³ + y² + 2z² substitiute x terms in for y and z

x(x+2)(x+4) = x³ + (x+2)² + 2(x+4)² solve for x by graphing on DEMOS

x = -2, 6 solved algebraically below

x = -2 y=0 z=2

x = 6 y=8 z=10

Checked both answers

xyz = x³ + y² + 2z²

-2(0)2 = -8 + 0 + 8 when x = -2

0 = 0

and

6(8)(10) = 6³ + 8² +2(10)² when x = 6

480 = 216+64+200

= 480

x(x+2)(x+4) = x³ + (x+2)² + 2(x+4)² solved algebraically

(x²+2x)(x+4) = x³ + x² +4x +4 + 2x² + 16x + 32

x³ + 6x² + 8x = x³ + x² +4x +4 + 2x² + 16x + 32

x³ + 6x² + 8x = x³ + 3x² + 20x + 36

3x² - 12x - 36 = 0 factor out the 3

3[x² - 4x - 12] = 0

3(x+2)(x-6) = 0 x = -2 and -12