Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given expression is

[tex]\sf \: {x}^{3} - \dfrac{1}{ {y}^{3} } - \dfrac{ {3x}^{2} }{y} + \dfrac{3x}{ {y}^{2} } \\ \\ [/tex]

can be rewritten as

[tex]\sf \: = \: {x}^{3} - {\bigg(\dfrac{1}{y} \bigg) }^{3} - 3 \times {x}^{2} \times \dfrac{1}{y} + 3 \times x \times \dfrac{1}{ {y}^{2} } \\ \\ [/tex]

can be rewritten as

[tex]\sf \: = \: {x}^{3} - 3 \times {x}^{2} \times \dfrac{1}{y} + 3 \times x \times \dfrac{1}{ {y}^{2} } - {\bigg(\dfrac{1}{y} \bigg) }^{3} \\ \\ [/tex]

We know,

[tex]\boxed{ \sf{ \: {a}^{3} - {3a}^{2}b + {3ab}^{2} - {b}^{3} = {(a - b)}^{3} \: }} \\ \\ [/tex]

So, here

[tex]\sf \: a = x \\ \\ [/tex]

[tex]\sf \: b = \dfrac{1}{y} \\ \\ [/tex]

So, on substituting the values, we get

[tex]\sf \: = \: {\bigg(x - \dfrac{1}{y} \bigg) }^{3} \\ \\ [/tex]

[tex]\sf \: = \: {\bigg(x - \dfrac{1}{y} \bigg) }{\bigg(x - \dfrac{1}{y} \bigg) }{\bigg(x - \dfrac{1}{y} \bigg) } \\ \\ [/tex]

Hence,

[tex]\sf \: \bf\implies \: {x}^{3} - \dfrac{1}{ {y}^{3} } - \dfrac{ {3x}^{2} }{y} + \dfrac{3x}{ {y}^{2} } \\ \\ \bf \: = \: {\bigg(x - \dfrac{1}{y} \bigg) }{\bigg(x - \dfrac{1}{y} \bigg) }{\bigg(x - \dfrac{1}{y} \bigg) } \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]