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We will now extend this idea of expanding algebraic terms/numbers with cubic order using the identities.
Let us derive the cubic identities with the help of known square identities.
1. We will now prove (a+b)^3= a^3+3a^2b+3ab^2+b^3 by direct multiplication.
Consider the LHS (a+b)^3.
Here (a+b) raised to the power 3. It means we need to multiply (a+b) by itself for two times.
That is (a+b)\times(a+b)\times(a+b) = (a+b)^3
(a+b)^3 = [(a+b)\times(a+b)]\times(a+b)
= (a+b)^2\times(a+b)
= (a^2+b^2+2ab)\times(a+b)
Apply the distributive property.
= a^2\times a+b^2\times a+2ab\times a+a^2\times b+b^2\times b+2ab\times b
= a^3+ab^2+2a^2b+a^2b+b^3+2ab^2
= a^3+3a^2b+3ab^2+b^3
= RHS
Thus, the identity is (a+b)^3 = a^3+3a^2b+3ab^2+b^3.
2. We will now prove (a-b)^3 = a^3-3a^2b+3ab^2-b^3 by direct multiplication.
Consider the LHS (a-b)^3.
Here (a-b) raised to the power 3. It means we need to multiply (a-b) by itself for two times.
That is (a-b)\times (a-b)\times(a-b)=(a-b)^3.
(a-b)^3 = [(a-b)\times(a-b)]\times(a-b)
= (a-b)^2\times(a+b)
= a^2+b^2-2ab\times(a-b)
Apply the distributive property.
= a^2\times a+(b^2\times a)+(-2ab\times a)+(a^2\times -b)b^2\times -b)+(-2ab\times -b)
= a^3+ab^2-2a^2b-a^2b-b^3+2ab^2
= a^3-3a^2b+3ab^2-b^3
= RHS
Thus, the identity is (a-b)^3 = a^3-3a^2b+3ab^2-b^3
3. We will now prove (x+a)(x+b)(x+c) = x^3+(a+b+c)x^2+(ab+bc+ca)x+abc.
Consider the LHS, (x+a)(x+b)(x+c).
Apply the distributive property for the first two terms.
[(x+a)(x+b)](x+c) = [(x\times x)]+(x\times b)+(a\times x)+(a\times b)](x+c)
= (x^2+bx+ax+ab)(x+c)
Again apply distributive law.
(x^2+bx+ax+ab)(x+c)
= (x^2\times x)+(bx\times x)+(ax\times x)+(ab\times x)+(x^2\times c)+(bx\times c)+(ax\times c)+(ab\times c).
= x^3+bx^2+ax^2+abx+cx^2+bcx+acx+abc
Separate the cubic, square, variables and constants terms.
= x^3+ax^2+bx^2+cx^2+abx+bcx+acx+abc
= x^3+(a+b+c)x^2+(ab+bc+ac)x+abc
= RHS
Thus we have the identity (x+a)(x+b)(x+c) = x^3+(a+b+c)x^2+(ab+bc+ca)x+abc.
Let us summarize identities.
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