A little help with algebra (homework)

I was brushing up on my algebra a little, and I've seen that I've become rusty with some of the concepts. Indications for solving these would perhaps be better suited rather than full answers, as the goal is to figure it out myself.

1. Show that for any $a, b \in \mathbb{R}$ (without induction, just using algebra) that $a^3 - b^3 = (a - b)(a^2 + 2ab + b^2)$ and something similar but for $a^3 + b^3$.

2. Factorize the following: $(3x + 1)^2 - (x+3)^2$ and other such expressions

And if someone could provide a concise and simple reference for the rules of expansions, factorizing and related topic, I would be most appreciative.

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Wouldn't just multiplying it out answer the question? –  John Smith Aug 20 '12 at 17:28

5 Answers

The simplest way to verify an identity like $a^3-b^3=(a-b)(a^2+2ab+b^2)$ is to multiply out the righthand side and verify that you do indeed get the lefthand side. If you try it in this case, however, you’ll fail:

\begin{align*} (a-b)(a^2+2ab+b^2)&=a(a^2+2ab+b^2)-b(a^2+2ab+b^2)\\ &=\left(a^3+2a^2b+ab^2\right)-\left(a^2b+2ab^2+b^3\right)\\ &=a^3+2a^2b+ab^2-a^2b-2ab^2-b^3\\ &=a^3-b^3+\left(2a^2b-a^2b\right)+\left(ab^2-2ab^2\right)\\ &=a^3-b^3+a^2b-ab^2\;, \end{align*}

and $a^2b-ab^2=ab(a-b)$ certainly isn’t guaranteed to be zero. (It’s zero if and only if either $a=0,b=0$, or $a=b$.) Thus, in general $a^3-b^3\ne(a-b)(a^2+2ab+b^2)$. The correct identity is $$a^3-b^3=(a-b)(a^2+ab+b^2)\;,$$ as you can check by multiplying out the righthand side: this time everything will cancel out except $a^3-b^3$.

The trick to factorizing an expression like $(3x + 1)^2 - (x+3)^2$ is to recognize that it has the form $a^2-b^2$, where $a=3x+1$ and $b=x+3$, and to recall the basic factorization formula $$a^2-b^2=(a-b)(a+b)\;.$$

A few of the standard basic formulas can be found here, together with a link to a practice page. Here is the start of a set of three pages on the topic, with examples. Googling on factoring formulas, with or without quotes, will turn up many more such resources.

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The first one should be $a^3-b^3=(a-b)(a^2+ab+b^2)$, and you can prove it most easily by simply expanding the right-hand side. Hover over the empty space below to see how it should turn out, after you try it yourself:

$(a-b)(a^2+ab+b^2)=a(a^2+ab+b^2)-b(a^2+ab+b^2)=a^3+a^2b+ab^2-a^2b-ab^2-b^3=a^3-b^3.$

For the $a^3+b^3$ identity, note that we can rewrite it as $a^3-(-b^3)=a^3-(-b)^3$ (since $3$ is odd). Thus, we can go into the right side of the first identity and plug in $(-b)$ anywhere we see $b$, and simplify, to get $a^3+b^3=\text{something factored}$. After you've worked it out yourself and checked your answer (both of which I strongly recommend for your own good), hover your mouse over the empty space below to see how it should turn out:

$a^3+b^3=a^3-(-b)^3=(a-(-b))(a^2+a(-b)+(-b)^2)=(a+b)(a^2-ab+b^2).$

For the second one, you should use the identity $a^2-b^2=(a+b)(a-b)$ to get your factored form. What should $a$ be in this case? What about $b$?

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For #1, just perform the multiplication on the right hand side of the "=" and you will have the left hand side.

For #2, since the two terms have no common factors you will have to multiply the expresion out and combine like terms. Then you will have a degree 4 polynomial that you need to apply various factorization strategies on.

If the polynomial only has $x^2$ and $x^4$, then you can use the quadratic formula on it to find solutions for $x^2$.

If there are $x$'s or $x^3$'s present, you will likely have to look for roots of the polynomial, and divide by $x-c$ for each root $c$ to factor out $x-c$. It is entirely possible that some of the roots are complex, so you should keep an eye out for that.

Update us on your work in progress.

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One possible approach is through using the binomial theorem: $$a^3 + b^3 = a^3 + 3a^2b + 3b^2a + b^3 - 3(a^2b + b^2a)$$ $$=(a+b)^3 - 3ab(a+b) = (a+b)((a+b)^2 - 3ab)$$ $$=(a+b)(a^2 - ab + b^2)$$ To derive the other sign, simply replace $b$ with $-b$ and you readily get $$a^3-b^3 = a^3 + (-b)^3 = (a-b)(a^2 + ab + b^2)$$

To solve number two, try a difference of squares which is given by the rule $$a^2 - b^2 = (a+b)(a-b)$$

As a note, the factoring in number 1 is called a difference (sum) of cubes, which is a special case of a very powerful technique for factoring called difference of powers (I suggest you google this if you wish to learn more) which is quite common in mathematics. The difference of squares in number 2 is also a special case of this technique.

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$1.$ Calculate $(a-b)(a^2+ab+b^2)$. This is $$a(a^2+ab+b^2)-b(a^2+ab+b^2).$$ You will after some simplification get $a^3-b^3$. Please note that you wrote $a^2+2ab+b^2$, which is not right. So it is not surprising that things were not working out!

For the second part, in $(1)$, replace $b$ everywhere by $-b$, and simplify a bit.

$2.$ Verify the important general identity $a^2-b^2=(a-b)(a+b)$. Then let $a=3x+1$ and $b=x+3$, and simplify a bit.

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