Question

Find three consecutive odd numbers the product of which is 148 more than the cube of their middle number.

Answer 1

Answer:

   The numbers are $-39,-37,-35$

Step-by-step explanation:

Let

x-----> the first odd number

x+2----> the middle odd number

x+4---> the third odd number

$x(x+2)(x+4)=(x+2)^{3} +148$

$(x^{2}+2x)(x+4)=(x^{2}+4x+4)(x+2)+148\\x^{3}+4x^{2}+2x^{2}+8x=x^{3}+2x^{2}+4x^{2}+8x+4x+8+148\\x^{3}+6x^{2}+8x=x^{3}+6x^{2}+12x+156\\8x=12x+156\\4x=-156\\x=-39$

The numbers are

$x=-39$

$x+2=-39+2=-37$

$x+4=-39+4=-35$