Question

factor this difference of cubes.\\(x^3 - 27\\)\\((x + ?)(x^2 + x + )\\)\\(\text{hint: } a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\\)

Explanation:

Step1: Identify \(a\) and \(b\)

Given the expression \(x^3 - 27\), we can rewrite it as \(x^3 - 3^3\). So, \(a = x\) and \(b = 3\) (since the formula for difference of cubes is \(a^3 - b^3=(a - b)(a^2+ab + b^2)\)).

Step2: Apply the difference of cubes formula

Using the formula \(a^3 - b^3=(a - b)(a^2+ab + b^2)\), substitute \(a = x\) and \(b = 3\).
First, the first factor is \((a - b)=(x - 3)\), but in the given form \((x + [?])\), we note that \((x - 3)=(x+(- 3))\), so the first box (the green one) is \(-3\).
Then, the second factor: \(a^2=x^2\), \(ab = x\times3 = 3x\), but in the formula for difference of cubes, the middle term is \(+ab\) (wait, no, correction: the formula is \(a^3 - b^3=(a - b)(a^2+ab + b^2)\)? Wait no, wait the hint says \(a^3\pm b^3=(a\pm b)(a^2\mp ab + b^2)\). Oh right! For difference of cubes (\(a^3 - b^3\)), it's \((a - b)(a^2+ab + b^2)\)? Wait no, the hint has a typo? Wait no, let's check the hint again: \(a^3\pm b^3=(a\pm b)(a^2\mp ab + b^2)\). So for \(a^3 - b^3\) (minus in the cube), it's \((a - b)(a^2+ab + b^2)\)? Wait no, the second sign: if the first is minus (\(a^3 - b^3\)), then the binomial is \((a - b)\) and the trinomial is \(a^2+ab + b^2\)? Wait no, let's take an example. Let \(a = 2\), \(b = 1\). \(2^3-1^3 = 8 - 1=7\). \((2 - 1)(4 + 2+1)=1\times7 = 7\). Correct. Another example: \(x^3 - 8=x^3 - 2^3=(x - 2)(x^2+2x + 4)\). So yes, the formula for \(a^3 - b^3=(a - b)(a^2+ab + b^2)\). But in the given problem, the binomial is written as \((x + [?])\), so we need to write \((x - 3)\) as \((x+(-3))\), so the first box is \(-3\). Then the trinomial: \(a^2=x^2\), \(ab=x\times3 = 3x\), \(b^2 = 9\). So the trinomial is \(x^2+3x + 9\)? Wait but the given trinomial is \((x^2+[?]x+[?])\). Wait, no, wait the user's problem has the trinomial as \(x^2+[?]x+[?]\). Wait, maybe I misread the hint. Wait the hint says \(a^3\pm b^3=(a\pm b)(a^2\mp ab + b^2)\). Oh! So for \(a^3 - b^3\) (the minus case), it's \((a - b)(a^2+ab + b^2)\)? No, wait the \(\mp\) in the trinomial. So if the binomial is \((a - b)\) (for \(a^3 - b^3\)), then the trinomial is \(a^2+ab + b^2\)? Wait no, let's check with \(a = x\), \(b = 3\), \(a^3 - b^3=x^3 - 27\). Using the formula from the hint: \(a^3 - b^3=(a - b)(a^2+ab + b^2)\)? Wait the hint's formula: \(a^3\pm b^3=(a\pm b)(a^2\mp ab + b^2)\). So for \(a^3 - b^3\) (the minus sign in the cube), the binomial is \((a - b)\) (so \(\pm\) is minus), and the trinomial is \(a^2\mp ab + b^2\), so \(\mp\) is plus (because \(\pm\) is minus, so \(\mp\) is plus). So \(a^2+ab + b^2\). So \(x^2+3x + 9\). But the given trinomial in the problem is \((x^2+[?]x+[?])\). Wait, but the first factor is \((x + [?])\), so we have to write \((x - 3)=(x+(-3))\), so the first box (green) is \(-3\), then the trinomial: \(x^2+3x + 9\)? Wait no, let's do it step by step.

Given \(x^3 - 27=x^3 - 3^3\).

Using the difference of cubes formula: \(a^3 - b^3=(a - b)(a^2+ab + b^2)\), where \(a = x\), \(b = 3\).

So \((a - b)=(x - 3)=(x+(-3))\), so the first box (the green one) is \(-3\).

Then the second factor: \(a^2=x^2\), \(ab=x\times3 = 3x\), \(b^2 = 9\). So the trinomial is \(x^2+3x + 9\)? Wait but the problem's trinomial is written as \((x^2+[?]x+[?])\). Wait, maybe the problem has a typo in the sign, but according to the formula, let's proceed.

Wait, the user's problem shows \((x + [?])(x^2+[?]x+[?])\). So we need to express \(x^3 - 27\) as \((x + c)(x^2+dx + e)\). Let's expand \((x + c)(x^2+dx + e)=x^3+dx^2+ex+cx^2+cdx+ce=x^3+(d + c)x^2+(e + cd)x+ce\). And we know that \(x^3 - 27=x^3+0x^2+0x - 27\). So we have the system of equations:

1. \(d + c=0\) (coefficient of \(x^2\))
2. \(e + cd=0\) (coefficient of \(x\))
3. \(ce=-27\) (constant term)

We also know from the difference of cubes that \(c=-3\) (since \(x^3 - 3^3=(x - 3)(x^2+3x + 9)\), so \(c=-3\)). Let's check:

If \(c=-3\), then from equation 3: \(-3e=-27\implies e = 9\).

From equation 1: \(d + (-3)=0\implies d = 3\).

Check equation 2: \(e + cd=9+(-3)\times3=9 - 9=0\), which matches.

So the first box (green) is \(-3\), the middle box (the coefficient of \(x\)) is \(3\), and the last box is \(9\). But the problem asks for the green box first, then the others, but let's focus on the green box first. Wait, the green box is in \((x + [?])\), so that's \(x+(-3)\), so the green box is \(-3\).

Answer:

The value in the green box (the first question mark) is \(-3\), the middle box (coefficient of \(x\)) is \(3\), and the last box is \(9\). But since the question seems to ask for the green box first, the answer for the green box is \(-3\).