Answer:
[tex]\boxed{ \bf\: {27y}^{3} + {125z}^{3} = (3y + 5z)\left({9y}^{2} - 15yz + {25z}^{2}\right) \: } \\ [/tex]
Step-by-step explanation:
Given algebraic expression is
[tex]\rm \: {27y}^{3} + {125z}^{3} \\ [/tex]
can be rewritten as
[tex]\rm \: = \: {(3y)}^{3} + {(5z)}^{3} \\ [/tex]
We know,
[tex]\boxed{ \sf\: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} ) \: } \\ [/tex]
So, using this algebraic identity, we get
[tex]\rm \: = \:(3y + 5z)[{(3y)}^{2} - (3y)(5z) + {(5z)}^{2}] \\ [/tex]
[tex]\rm \: = \:(3y + 5z)\left({9y}^{2} - 15yz + {25z}^{2}\right) \\ [/tex]
Hence,
[tex]\implies\boxed{ \bf\: {27y}^{3} + {125z}^{3} = (3y + 5z)\left({9y}^{2} - 15yz + {25z}^{2}\right) \: } \\ [/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]
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Answers & Comments
Answer:
[tex]\boxed{ \bf\: {27y}^{3} + {125z}^{3} = (3y + 5z)\left({9y}^{2} - 15yz + {25z}^{2}\right) \: } \\ [/tex]
Step-by-step explanation:
Given algebraic expression is
[tex]\rm \: {27y}^{3} + {125z}^{3} \\ [/tex]
can be rewritten as
[tex]\rm \: = \: {(3y)}^{3} + {(5z)}^{3} \\ [/tex]
We know,
[tex]\boxed{ \sf\: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} ) \: } \\ [/tex]
So, using this algebraic identity, we get
[tex]\rm \: = \:(3y + 5z)[{(3y)}^{2} - (3y)(5z) + {(5z)}^{2}] \\ [/tex]
[tex]\rm \: = \:(3y + 5z)\left({9y}^{2} - 15yz + {25z}^{2}\right) \\ [/tex]
Hence,
[tex]\implies\boxed{ \bf\: {27y}^{3} + {125z}^{3} = (3y + 5z)\left({9y}^{2} - 15yz + {25z}^{2}\right) \: } \\ [/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]