Question

Prove the following
`\bold{algebraically}`:


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

Answer 1

Explanation:

Identity to use:

1+N+N^2+N^3 = (N^4-1)/(N-1)

Let N=29

1+29+29^2+29^3 = (29^4-1) / (29-1)

30+29^2+29^3 = (29^4-1) / 28

Transpose and re-arrange

(29^3+29^2+30) / (29^4-1) = 1 / 28 QED

Answer 2

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