UPSKILL MATH PLUS
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Learn moreWe 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 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 \((a+b)^3\) \(=\) \(a^3+3a^2b+3ab^2+b^3\).
Take the factor \(3ab\) from the middle two terms of RHS we get:
\((a+b)^3\) \(=\) \(a^3+b^3+3ab(a+b)\)
2. We will now prove \((a-b)^3\) \(=\) \(a^3-3a^2b+3ab^2-b^3\) by replacing \(b\) by \(-b\) in the above identity.
Consider the identity \((a+b)^3\) \(=\) \(a^3+3a^2b+3ab^2+b^3\).
Replace \(b\) by \(-b\) in the above equation.
\((a-b)^3\) \(=\) \(a^3+3a^2(-b)+3a(-b)^2+(-b)^3\)
\(=\) \(a^3-3a^2b+3ab^2-b^3\)
Thus, the identity is \((a-b)^3\) \(=\) \(a^3-3a^2b+3ab^2-b^3\)
Take the factor \(-3ab\) from the middle two terms of RHS we get:
\((a-b)^3\) \(=\) \(a^3+b^3-3ab(a-b)\)
Let us summarize the identities...
Identity VII: \((a-b)^3\) \(=\) \(a^3-3a^2b+3ab^2-b^3\) or \((a-b)^3\) \(=\) \(a^3-b^3-3ab(a-b)\)
Important!
The formula for three different values:
The general form of the above two identities is given by the following identity expression:
\((x+a)(x+b)(x+c)\) \(=\) \(x^3\)\(+(a+b+c)x^2\)\(+(ab+bc+ca)x+abc\).
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 \((x+a)(x+b)(x+c)\) \(=\) \(x^3\)\(+(a+b+c)x^2\)\(+(ab+bc+ca)x+abc\).
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