Question

`\tiny \displaystyle \bf \red{\int^(3)_(1) \int^(3)_(1) \int^(3)_(1) \int^(3)_(1) ( \alpha + \beta + \gamma + \theta - \eta)/(\alpha + \beta + \gamma + \theta + \eta) \: d \alpha d \beta d \gamma d \theta d \eta}`

Cʜᴀʟʟᴇɴɢᴇ❤️❤️❤️❤️❤️​

Answer 1

I = 96/10 = 96:10 is your correct answer.

Answer 2

Explanation:

`\displaystyle \sf \red{I=\int^(3)_(1) \int^(3)_(1) \int^(3)_(1) \int^(3)_(1) ( \alpha + \beta + \gamma + \theta - \eta)/(\alpha + \beta + \gamma + \theta + \eta) \: d \alpha d \beta d \gamma d \theta d \eta}`

According to symmetry

`\displaystyle \sf \red{I=\int^(3)_(1) \int^(3)_(1) \int^(3)_(1) \int^(3)_(1) ( \alpha + \beta + \gamma - \theta + \eta)/(\alpha + \beta + \gamma + \theta + \eta) \: d \alpha d \beta d \gamma d \theta d \eta}`

`\displaystyle \sf \red{I=\int^(3)_(1) \int^(3)_(1) \int^(3)_(1) \int^(3)_(1) ( \alpha + \beta - \gamma + \theta + \eta)/(\alpha + \beta + \gamma + \theta + \eta) \: d \alpha d \beta d \gamma d \theta d \eta}`

`\displaystyle \sf \red{I=\int^(3)_(1) \int^(3)_(1) \int^(3)_(1) \int^(3)_(1) ( \alpha - \beta + \gamma + \theta + \eta)/(\alpha + \beta + \gamma + \theta + \eta) \: d \alpha d \beta d \gamma d \theta d \eta}`

`\displaystyle \sf \red{I=\int^(3)_(1) \int^(3)_(1) \int^(3)_(1) \int^(3)_(1) ( - \alpha + \beta + \gamma + \theta + \eta)/(\alpha + \beta + \gamma + \theta + \eta) \: d \alpha d \beta d \gamma d \theta d \eta}`

Take sum of 5 integrals

`\displaystyle \sf \green{5I=\int^(3)_(1) \int^(3)_(1) \int^(3)_(1) \int^(3)_(1) ( 3(\alpha + \beta + \gamma + \theta + \eta))/(\alpha + \beta + \gamma + \theta + \eta) \: d \alpha d \beta d \gamma d \theta d \eta}`

`\displaystyle \sf \pink{5I=3\int^(3)_(1) \int^(3)_(1) \int^(3)_(1) \int^(3)_(1) \: d \alpha d \beta d \gamma d \theta d \eta}</p><p>`

`\large \sf \green{5I=3( {2)}^(5) }`

`\large \sf \pink{I={ (96)/(5) }^{} }`