Question

Prove the following
`\bold{algebraically}`:


`\displaystyle \frac{ {29}^(3) + {29}^(2) + 30 }{ {29}^(4) - 1} = (1)/(28)`
​​

Answer 1

see below

Explanation:

we are given

`\displaystyle \frac{ {29}^(3) + {29}^(2) + 30 }{ {29}^(4) - 1} = (1)/(28)`

we want to prove it algebraically

to do so rewrite 30:

`\displaystyle \frac{ {29}^(3) + {29}^(2) + 29 + 1}{ {29}^(4) - 1} \stackrel{ ? }{= }(1)/(28)`

let 29 be a thus substitute:

`\displaystyle \frac{ {a}^(3) + {a}^(2) + a + 1}{ {a}^(4) - 1} \stackrel{ ? }{= }(1)/(28)`

factor the denominator:

`\rm\displaystyle \frac{ {a}^(3) + {a}^(2) + a + 1}{ ({a}^(2) + 1) (a- 1)(a + 1)} \stackrel{ ? }{= }(1)/(28)`

Factor out a²:

`\rm\displaystyle \frac{ {a}^(2) ({a}^{} + 1)+ a + 1}{ ({a}^(2) + 1) (a- 1)(a + 1)} \stackrel{ ? }{= }(1)/(28)`

factor out 1:

`\rm\displaystyle \frac{ {a}^(2) ({a}^{} + 1)+1( a + 1)}{ ({a}^(2) + 1) (a- 1)(a + 1)} \stackrel{ ? }{= }(1)/(28)`

group:

`\rm\displaystyle \frac{ ({a}^(2) +1)( a + 1)}{ ({a}^(2) + 1) (a + 1)(a - 1)} \stackrel{ ? }{= }(1)/(28)`

reduce fraction:

`\rm\displaystyle \frac{ \cancel{({a}^(2) +1)( a + 1)}}{ \cancel{({a}^(2) + 1) (a + 1)}(a - 1)} \stackrel{ ? }{= }(1)/(28)`

`\displaystyle (1)/(a - 1) \stackrel {?}{ = } (1)/(28)`

substitute back:

`\displaystyle (1)/(29 - 1) \stackrel {?}{ = } (1)/(28)`

simplify substraction:

`\displaystyle (1)/(28) \stackrel { \checkmark}{ = } (1)/(28)`

hence Proven

Answer 2

Solution given:

L.H.S.

`\displaystyle \frac{ {29}^(3) + {29}^(2) + 29+1 }{ {29}^(4) - 1}`

Let 29 be x.

`\displaystyle \frac{ {x}^(3) + {x}^(2) + x+1 }{ {x}^(4) - 1}`

`\displaystyle \frac{ {x}^(2) (x+ 1) +1( x+1 )}{ ({x}^(2) )²- 1²}`

`\displaystyle ( (x²+1)( x+1 ))/( (x²+1)(x²-1²))`

`\displaystyle ( (x²+1)( x+1 ))/( (x²+1)(x+1)(x-1))`

`\displaystyle (1)/(x-1)`

substituting value.

`\displaystyle (1)/(29-1)`

`\displaystyle (1)/(28)`

R.H.S.

proved.