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Theory:

Let us derive the cubic identities with the help of known identitites.
Expansion of \((x+y)^3\):
Substitute \(a = b = c = y\) in the identity \((x+a)(x+b)(x+c)\) \(=\) \(x^3\)\(+(a+b+c)x^2\)\(+(ab+bc+ca)x+abc\).
 
Consider the LHS, \((x+a)(x+b)(x+c)\).
 
\((x+a)(x+b)(x+c)\) \(=\) \((x+y)(x+y)(x+y)\)
 
\(=\) \((x+y)^{3}\)
 
Consider the RHS, \(x^3\)\(+(a+b+c)x^2\)\(+(ab+bc+ca)x+abc\).
 
\(x^3\)\(+(a+b+c)x^2\)\(+(ab+bc+ca)x+abc\) \(=\) \(x^3\)\(+(y+y+y)x^2\)\(+(yy+yy+yy)x+yyy\)
 
\(=\) \(x^3\)\(+(3y)x^2\)\(+(y^2+y^2+y^2)x+y^3\)
 
\(=\) \(x^3\)\(+3yx^2\)\(+3y^2x+y^3\)
 
Thus, the identity is \((x+y)^3\) \(=\) \(x^3+3x^2y+3xy^2+y^3\).
 
The obtained cubic identity can also be rewritten as follows:
 
Consider the standard identity, \((x+y)^3\)\(=\)\(x^3+3x^2y\)\(+3xy^2+y^3\).
 
Take the factor \(3xy\) from the middle two terms of RHS.
 
Thus, x+y3=x3+y3+3xyx+y.
Expansion of \((x-y)^3\):
Replace \(y\) by \(-y\) in the cubic identity of \((x+y)^3\)\(=\)\(x^3+3x^2y\)\(+3xy^2+y^3\).
 
\((x+(-y))^3\) \(=\) \(x^3+3x^2(-y)+3x(-y)^2+(-y)^3\)
 
\((x-y)^3\) \(=\) \(x^3-3x^2y+3xy^2-y^3\)
 
Thus, the identity is \((x-y)^3\) \(=\) \(x^3-3x^2y+3xy^2-y^3\)
 
The obtained cubic identity can also be rewritten as follows:
 
Consider the standard identity, \((x-y)^3\) \(=\) \(x^3-3x^2y+3xy^2-y^3\).
 
Take the factor \(3xy\) from the middle two terms of RHS.
 
Thus, xy3=x3y3+3xyxy
 
Let us summarize the identities...
 
  • \((x+y)^3\)\(=\)\(x^3+3x^2y\)\(+3xy^2+y^3\) or x+y3=x3+y3+3xyx+y
  • \((x-y)^3\) \(=\) \(x^3-3x^2y+3xy^2-y^3\) or xy3=x3y3+3xyxy
 
Click here! to explore some examples on the expansion of cubic terms.