Question

`\sf \bf \orange{Prove \: \bf \red{That:}}`


`\sf\left((x^(a^2+b^2))/(x^(ab))\right)^(a+b)\left((x^(b^2+c^2))/(x^(bc))\right)^(b+c)\left((x^(c^2+a^2))/(x^(ca))\right)^(c+a)=x^(2\left(a^3+b^3+c^3\right))`

No Spam:)

︎
︎
︎
︎
︎
︎
︎
︎
︎
︎
Thank You!​

Answer 1

`\sf\left((x^(a^2+b^2))/(x^(ab))\right)^(a+b)\left((x^(b^2+c^2))/(x^(bc))\right)^(b+c)\left((x^(c^2+a^2))/(x^(ca))\right)^(c+a) \\ = {x}^{ ({a}^(2) + {b}^(2) - ab)(a + b)} \: \: {x}^{ ({b}^(2) + {c}^(2) - bc)( b + c)} \: \: {x}^{ ({c}^(2) + {a}^(2) - ca)(c + a)} \\ = {x}^{ {a}^(3) + {b}^(3 ) - {a}^(2)b + {a}^(2) b + {b}^(3) - a {b}^(2) } \: \: {x}^{ {b}^(3) + {c}^(3 ) - {b}^(2)c + {b}^(2) c + {c}^(3) - b{c}^(2) } \: \: {x}^{ {c}^(3) + {a}^(3 ) - {c}^(2)a + {c}^(2) a + {c}^(3) - c {a}^(2) } \: \: \\ = {x}^{{a}^(3) + {b}^(3 ) - {a}^(2)b + {a}^(2) b + {b}^(3) - a {b}^(2) + {b}^(3) + {c}^(3 ) - {b}^(2)c + {b}^(2) c + {c}^(3) - b{c}^(2) + {c}^(3) + {a}^(3 ) - {c}^(2)a + {c}^(2) a + {c}^(3) - c {a}^(2)} \\ </p><p> = {x}^{ {a}^(3) + {b}^(3) } \: \: {x}^{ {b}^(3) + {c}^(3) } \: \: {x}^{ {c}^(3) + {a}^(3) } \\ = {x}^{{a}^(3) + {b}^(3) + {b}^(3) + {c}^(3) + {c}^(3) + {a}^(3)} </p><p> \\ = {x}^{2 {a}^(3) + 2 {b}^(3) + 2 {c}^(3) } \\ = {x}^{2( {a}^(3) + {b}^(3) + {c}^(3) }`

Hope you could get an idea from here.

Doubt clarification - use comment section.