Verified answer

[tex]\large\underline{\sf{Solution-1}}[/tex]

Given that,

[tex]\sf \: x + y = 5 - - - (1) \\ [/tex]

and

[tex]\sf \: x - y = 3 - - - (2) \\ [/tex]

On adding equation (1) and (2), we get

[tex]\sf \: 2x = 8 \: \: \implies\sf \: x = 4 \\ [/tex]

On substituting x = 4 in equation (1), we get

[tex]\implies\sf \: y = 1 \\ [/tex]

Now, Consider

[tex]\sf \: {x}^{3} + {y}^{3} + 20xy \\ [/tex]

[tex]\sf \: = \: {(4)}^{3} + {(1)}^{3} + 20(4)(1) \\ [/tex]

[tex]\sf \: = \:64 + 1 + 80 \\ [/tex]

[tex]\sf \: = \:145 \\ [/tex]

Hence,

[tex]\implies\sf \: \sf \: \boxed{\sf \: {x}^{3} + {y}^{3} + 20xy = 145 \: } \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-2}}[/tex]

Given that,

[tex]\sf \: a + \dfrac{1}{a} = 2 \\ [/tex]

[tex]\sf \: \dfrac{ {a}^{2} + 1}{a} = 2 \\ [/tex]

[tex]\sf \: { {a}^{2} + 1} = 2a \\ [/tex]

[tex]\sf \: { {a}^{2} + 1} - 2a = 0\\ [/tex]

[tex]\sf \: {(a - 1)}^{2} = 0\\ [/tex]

[tex]\sf \: a - 1 = 0 \\ [/tex]

[tex]\implies\sf \: \sf \: a =1 \\ [/tex]

Now, Consider

[tex]\sf \: {a}^{5} + \dfrac{1}{ {a}^{5} } \\ [/tex]

On substituting a = 1, we get

[tex]\sf \: = \: {(1)}^{5} + \dfrac{1}{ {(1)}^{5} } \\ [/tex]

[tex]\sf \: = \: 1 + 1 \\ [/tex]

[tex]\sf \: = \: 2 \\ [/tex]

Hence,

[tex]\implies\sf \: \sf \: \boxed{\sf \: {a}^{5} + \dfrac{1}{ {a}^{5} } = 2 \: } \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-3}}[/tex]

[tex]\sf \: {x}^{6} - {y}^{6} \\ [/tex]

[tex]\sf \: = \: {( {x}^{3}) }^{2} - {( {y}^{3} )}^{2} \\ [/tex]

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

[tex]\sf \: = \: (x - y)( {x}^{2} + xy + {y}^{2})(x + y)( {x}^{2} - xy + {y}^{2}) \\ [/tex]

Hence,

[tex]\boxed{\sf \: {x}^{6} - {y}^{6}= (x - y)( {x}^{2} + xy + {y}^{2})(x + y)( {x}^{2} - xy + {y}^{2}) \: } \\ \\ [/tex]