irene12 said:

Hello,I don't remember exactly how to finish

The product of three consecutive odd integers reduced by 23 is 99 less than the cube of the sum of the smallest number and 2. Compute the mean of the three integers.

(x)(x+2)(x+4)-23=(x+2)[sup:2jlksm1n]3[/sup:2jlksm1n]-99

(x)(x+4)-23=(x+2)2-99 >>>>>>>>>>>>>> Incorrect step

2x+4x=(x+2)2-76

6x=x2+4-76

x=x2/6-72/6

x=x2/6-12

x=x2-2

I don't think that i got it quite right?

(x)(x+2)(x+4)-23=(x+2)[sup:2jlksm1n]3[/sup:2jlksm1n]-99

(x)(x+2)(x+4) - (x+2)[sup:2jlksm1n]3[/sup:2jlksm1n] = 23 - 99

(x+2)[(x+2)[sup:2jlksm1n]2[/sup:2jlksm1n] - x(x+4)] = 76

(x+2)[x[sup:2jlksm1n]2[/sup:2jlksm1n] + 4x + 4 - x[sup:2jlksm1n]2[/sup:2jlksm1n] - 4x] = 76

Now continue....