Question

please help me doing this maths sum​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\: {\bigg(\dfrac{ {x}^{ {a}^{2} + {b}^{2} } }{ {x}^{ - ab} } \bigg) }^{a - b}$

$\red{\bigg \{ \because \: {x}^{m} \div {x}^{n} = {x}^{m - n} \bigg \}}$

$\rm \: = \: \: {\bigg( {x}^{ {a}^{2} + {b}^{2} + ab } \bigg) }^{a - b}$

$\rm \: = \: \: {\bigg({x}^{({a}^{2} + {b}^{2} + ab)(a - b)} \bigg) }$

$\rm \: = \: \: {x}^{ {a}^{3} - {b}^{3} }$

$\bf :\longmapsto\: {\bigg(\dfrac{ {x}^{ {a}^{2} + {b}^{2} } }{ {x}^{ - ab} } \bigg) }^{a - b} = {x}^{ {a}^{3} - {b}^{3} }$

Consider,

$\rm :\longmapsto\: {\bigg(\dfrac{ {x}^{ {b}^{2} + {c}^{2} } }{ {x}^{ - bc} } \bigg) }^{b - c}$

$\rm \: = \: \: {\bigg( {x}^{ {b}^{2} + {c}^{2} + bc } \bigg) }^{b - c}$

$\red{\bigg \{ \because \: {x}^{m} \div {x}^{n} = {x}^{m - n} \bigg \}}$

$\rm \: = \: \: {\bigg({x}^{({b}^{2} + {c}^{2} + bc)(b - c)} \bigg) }$

$\rm \: = \: \: {x}^{ {b}^{3} - {c}^{3} }$

$\bf :\longmapsto\: {\bigg(\dfrac{ {x}^{ {b}^{2} + {c}^{2} } }{ {x}^{ - bc} } \bigg) }^{b - c} = {x}^{ {b}^{3} - {c}^{3} }$

Consider,

$\rm :\longmapsto\: {\bigg(\dfrac{ {x}^{ {c}^{2} + {a}^{2} } }{ {x}^{ - ca} } \bigg) }^{c - a}$

$\rm \: = \: \: {\bigg( {x}^{ {c}^{2} + {a}^{2} + ac } \bigg) }^{c - a}$

$\red{\bigg \{ \because \: {x}^{m} \div {x}^{n} = {x}^{m - n} \bigg \}}$

$\rm \: = \: \: {\bigg({x}^{({c}^{2} + {a}^{2} + ac)(c - a)} \bigg) }$

$\rm \: = \: \: {x}^{ {c}^{3} - {a}^{3} }$

$\bf :\longmapsto\: {\bigg(\dfrac{ {x}^{ {c}^{2} + {a}^{2} } }{ {x}^{ - ca} } \bigg) }^{c - a} = {x}^{ {c}^{3} - {a}^{3} }$

Now, Consider

$\red{\rm :\longmapsto\:{\bigg(\dfrac{ {x}^{ {a}^{2} + {b}^{2} } }{ {x}^{ - ab} } \bigg) }^{a - b}{\bigg(\dfrac{ {x}^{ {b}^{2} + {c}^{2} } }{ {x}^{ - bc} } \bigg) }^{b - c}{\bigg(\dfrac{ {x}^{ {c}^{2} + {a}^{2} } }{ {x}^{ - ca} } \bigg) }^{c - a}}$

$\rm \: = \: \: {x}^{ {a}^{3} - {b}^{3} } \: \times \: {x}^{ {b}^{3} - {c}^{3} } \: \times \: {x}^{ {c}^{3} - {a}^{3} }$

$\rm \: = \: \: {x}^{ {a}^{3} - {b}^{3} + {b}^{3} - {c}^{3} + {c}^{3} - {a}^{3} } \:$

$\red{\bigg \{ \because \: {x}^{m} \times {x}^{n} = {x}^{m + n} \bigg \}}$

$\rm \: = \: \: \: {x}^{0}$

$\rm \: = \: \: \: 1$

Hence, Proved

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)