Question

Find the numbers a, b and c if the polynomial x3−18x2+24xb−c is the cube of the binomial x+2a.

Answer 1

`a=-3\\ \\b=(9)/(2)\\ \\c=216`

Explanation:

Use formula

`(u+v)^3=u^3+3u^2v+3uv^2+v^3.`

Hence,

`(x+2a)^3=x^3+3x^2\cdot 2a+3x\cdot (2a)^2+(2a)^3=x^3+6ax^2+12a^2x+8a^3.`

This expression is equal to

`x^3-18x^2+24bx-c,`

so we can equate the coefficients at powers of x:

`x^3:\ 1=1\\ \\x^2:\ 6a=-18\Rightarrow a=-3\\ \\x:\ 12a^2=24b\Rightarrow b=(9)/(2)\\ \\1=x^0:\ 8a^3=-c\Rightarrow c=-8\cdot (-3)^3=216.`