Theory:

Let us expand some of the cubic terms using its identities.
Write the cube in expanded form.
 
1. \((2x+3y)^3\)
 
Let us use the identity, \((a+b)^3\)\(=\) \(a^3+3a^2b+3ab^2+b^3\).
 
Comparing \((2x+3y)^3\) with \((a+b)^3\), we have \(a=2x\) and \(b=3y\).
 
Substitute the values in the formula.
 
\((2x+3y)^3\) \(=\) \((2x)^3\)\(+3(2x)^2(3y)\)\(+3(2x)(3y)^2\)\(+(3y)^3\)
 
\((2x+3y)^3\) \(=\) \(8x^3\)\(+(2\times 4\times 3)x^2y\)\(+(3\times 2\times 9)xy^2\)\(+27y^3\)
 
\(= 8x^3+24x^2y+54xy^2+27y^3\)
 
 
2. \((5x-7y)^3\)
 
Let us use the identity, \((a-b)^3\)\(=\)\(a^3-3a^2b+3ab^2-b^3\).
 
Comparing \((5x-7y)^3\) with \((a-b)^3\), we have \(a=5x\) an d\(b=7y\).
 
Substitute the values in the formula.
 
\((5x-7y)^3\) \(=\) \((5x)^3\)\(-3(5x)^2(7y)\)\(+3(5x)(7y)^2\)\(+(7y)^3\)
 
\((5x-7y)^3\) \(=\) \(125x^3\)\(-(3\times 25\times 7)x^2y\)\(+(3\times 5 \times 49)xy^2\)\(+343y^3\)
 
\((5x-7y)^3\) \(=\) \(125x^3\)\(-525x^2y\)\(+735xy^2\)\(+343y^3\)
 
 
4. \((4y+5)(4y+3)(4y-7)\)
 
Let us use the identity, \((x+a)(x+b)(x+c)\) \(=\) \(x^3+(a+b+c)x^2\)\(+(ab+bc+ca)x\)\(+abc\)
 
Comparing \((4y+5)(4y+3)(4y-7)\) with \((x+a)(x+b)(x+c)\), we have \(x=4y, a=5, b=3\) an d\(c=-7\).
 
Substitute the known values.
 
\((4y+5)(4y+3)(4y-7)\) \(=\) \((4y)^3\)\(+(5+3-7)(4y)^2\)\(+((5\times 3) + (3\times -7) +(-7\times 5))(4y)\)\(+5 \times 3 \times -7)\)
 
\((4y+5)(4y+3)(4y-7)\) \(=\) \(64y^3+16y^2\)\(+(15-21-35)(4y)-105\)
 
\((4y+5)(4y+3)(4y-7)\) \(=\) \(64y^3+16y^2\)\(-164y-105\)
 
Example:
Look for the following cases where we used the identities.
 
1. Expand \((y-5)^3\) using identity.
 
The above expression is in \((a-b)^3\) form.
 
We have the identity, \((a-b)^3\)\(=\)\(a^3-3a^2b+3ab^2-b^3\).
 
Substitute \(a = y\) and \(b = 5\) in the formula.
 
y53=y33(y)2(5)+3(y)(5)253
 
y53=y315y2+75y125
 
 
2. Evaluate \(103^3\) using identity.
 
\(103^3\) \(=\) \((100+3)^3\)
 
The above expression is in \((a+b)^3\) form.
 
We have the identity, \((a+b)^3\) \(=\) \(a^3+3a^2b+3ab^2+b^3\)
 
Substitute \(a =100\) and \(b = 3\) in the formula.
 
100+32=1003+3(100)2(3)+3(100)(3)2+33
 
=1000000+(3×10000×3)+(3×100×9)+27
 
=1000000+90000+2700+27
 
=1092727