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\begin{array}{rrcl} \text{Identity I:} & (a+b)^2 & = & a^2+b^2+2ab \\ \text{Identity II:} & (a – b)^2 & = & a^2 – 2ab + b^2 \\ \text{Identity III:} & a^2 – b^2 & = & (a + b)(a – b) \\ \text{Identity IV:} & (x + a)(x + b) & = & x^2 + (a + b) x + ab \\ \text{Identity V:} & (a + b + c)^2 & = & a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \\ \text{Identity VI:} & (a + b)^3 & = & a^3 + b^3 + 3ab (a + b) \\ \text{Identity VII:} & (a – b)^3 & = & a^3 – b^3 – 3ab (a – b) \\ \text{Identity VIII:} & a^3+b^3+c^3–3abc & = & (a+b+c)(a^2+b^2+c^2–ab–bc–c) \\ \end{array}

I have been studying this since my fifth grade ,I have also used this equations for many times in my problem . My question is on what basis this equation was derived?not only this equation there 8 equations for solving complicated problems. How were these 8 equations was this derived and is there any method for deriving it?I want to know it because I am curious and I want to make mathematics more easier for me to study .I am tried of studying it by these formula.

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    $\begingroup$ Where's the other seven? $\endgroup$ – Sean Roberson May 21 '18 at 7:54
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    $\begingroup$ All of them can be derived trough simple mutiplication. If you are interested in a general form look up the binomial theorem. $\endgroup$ – h3h325 May 21 '18 at 7:59
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    $\begingroup$ I have also used this equations for many times in my problem --- I find it hard to understand how you could have been using these equations before you knew how to multiply polynomials. When I began studying algebra, probably much later than you did (Junior High 8th grade by myself; High School 9th grade for course credit), multiplying polynomials by monomials, and multiplying polynomials by polynomials, were studied well before the topic of factoring, and both were covered well before they were used (such as in solving non-first degree equations and combining rational expressions). $\endgroup$ – Dave L. Renfro May 21 '18 at 8:11
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    $\begingroup$ Perhaps the following answers to other questions will help. The first answer is very basic, and the second answer builds on the first answer. How to teach algebraic expressions? AND How to expand $(a_0+a_1x+a_2x^2+...a_nx^n)^2$? $\endgroup$ – Dave L. Renfro May 21 '18 at 8:19
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    $\begingroup$ See here for how to use the factor theorem from precalculus math to find the right side of VIII when you only know the left side of VIII. For simpler factor theorem uses, if you need them, see here and here and here. $\endgroup$ – Dave L. Renfro May 21 '18 at 8:33
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This is just multiplying out the $(a+b)^{2}$. Observe that $$ (a+b)^{2} = (a+b)(a+b) = a(a+b) + b(a+b) = a^{2} + ab + ab + b^{2} = a^{2}+2ab+b^{2}. $$


For more general powers, you might know the binomial formula, which is $$ (a + b)^{n} = \sum_{i=0}^{n}\binom{n}{i}a^{n-i}b^{i} = a^{n} + na^{n-1}b + \cdots + nab^{n-1} + b^{n}. $$ If this looks scary at the moment, don't worry about it for now. To expand on Djhoe's comment about Pascal's triangle, the binomial coefficients (the numbers in front of the powers of $a$ and $b$) can be found in the following triangle: $$ \begin{array}{c} 1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \ \end{array} $$ Each number is the sum of the two above it. For example, the $6$ is the sum of the two $3$'s above it. These give you the coefficients of the powers of $(a + b)^{n}$: $$ \begin{array}{rc} (a + b)^{0} = & 1 \\ (a + b)^{1} = & a + b \\ (a + b)^{2} = & a^{2} + \color{red}{2}ab + b^{2} \\ (a + b)^{3} = & a^{3} + \color{red}{3}a^{2}b + \color{red}{3}ab^{2} + b^{3} \\ (a + b)^{4} = & a^{4} + \color{red}{4}a^{3}b + \color{red}{6}a^{2}b^{2} + \color{red}{4}ab^{3} + b^{4} \\ \end{array} $$ For the $0$ power, remember that anything to the power of $0$ is $1$.


All of the identities can be proved by following a similar method to the above. These are actually great exercises for you, so I recommend doing them yourself by expanding the brackets and simplifying. Here are a couple more examples for you to see the method and apply yourself:
Identity II is $$ (a - b)^{2} = (a - b)(a - b) = a(a - b) - b(a - b) = a^{2} - ab - ab + b^{2} = a^{2} - 2ab + b^{2}. $$ Identity V is $$ (a + b + c)^{2} = (a + b + c)(a + b + c) = a(a + b + c) + b(a + b + c) + c(a + b + c) $$ where we distribute the left bracket, and then this equals $$ (a^{2} + ab + ac) + (ab + b^{2} + bc) + (ac + bc + c^{2}) = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac $$ by collecting like terms.

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In order to evaluate $(a + b)^2$, we first need to think what the squaring means. It means multiplying the thing with itself. So now $$ (a+b)^2 = (a+b) \times (a+b) $$ Now comes the tricky part: We must open the parentheses. Let's start carefully and open the first parentheses first: $$ (a+b)^2 = (a+b) \times (a+b) = a\times (a+b) + b \times (a+b) $$ Here we used the distributive law of the real numbers (we always assume that $a$ and $b$ are real numbers). We can then open the second parentheses and again use the distributive law: $$ (a+b)^2 = a\times (a+b) + b \times (a+b) = a\times a + a\times b + b \times a + b\times b $$ For shorthand, let's take away the multiplication signs and remember the definition of squaring: $$ (a+b)^2 = a^2 + ab + ba + b^2 $$ We also remember that real numbers are commutative, that is $ab= ba$. Now we have two times the same term $ab$, so $$ (a+b)^2 = a^2 + 2ab + b^2 $$ If you completely understand this derivation, the rest of the identities should be easy to check.

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