Answer:
see below
Step-by-step explanation:
we are given
[tex] \displaystyle \frac{ {29}^{3} + {29}^{2} + 30 }{ {29}^{4} - 1} = \frac{1}{28}[/tex]
we want to prove it algebraically
to do so rewrite 30:
[tex] \displaystyle \frac{ {29}^{3} + {29}^{2} + 29 + 1}{ {29}^{4} - 1} \stackrel{ ? }{= }\frac{1}{28}[/tex]
let 29 be a thus substitute:
[tex] \displaystyle \frac{ {a}^{3} + {a}^{2} + a + 1}{ {a}^{4} - 1} \stackrel{ ? }{= }\frac{1}{28}[/tex]
factor the denominator:
[tex] \rm\displaystyle \frac{ {a}^{3} + {a}^{2} + a + 1}{ ({a}^{2} + 1) (a- 1)(a + 1)} \stackrel{ ? }{= }\frac{1}{28}[/tex]
Factor out a²:
[tex] \rm\displaystyle \frac{ {a}^{2} ({a}^{} + 1)+ a + 1}{ ({a}^{2} + 1) (a- 1)(a + 1)} \stackrel{ ? }{= }\frac{1}{28}[/tex]
factor out 1:
[tex] \rm\displaystyle \frac{ {a}^{2} ({a}^{} + 1)+1( a + 1)}{ ({a}^{2} + 1) (a- 1)(a + 1)} \stackrel{ ? }{= }\frac{1}{28}[/tex]
group:
[tex] \rm\displaystyle \frac{ ({a}^{2} +1)( a + 1)}{ ({a}^{2} + 1) (a + 1)(a - 1)} \stackrel{ ? }{= }\frac{1}{28}[/tex]
reduce fraction:
[tex] \rm\displaystyle \frac{ \cancel{({a}^{2} +1)( a + 1)}}{ \cancel{({a}^{2} + 1) (a + 1)}(a - 1)} \stackrel{ ? }{= }\frac{1}{28}[/tex]
[tex] \displaystyle \frac{1}{a - 1} \stackrel {?}{ = } \frac{1}{28} [/tex]
substitute back:
[tex] \displaystyle \frac{1}{29 - 1} \stackrel {?}{ = } \frac{1}{28} [/tex]
simplify substraction:
[tex] \displaystyle \frac{1}{28} \stackrel { \checkmark}{ = } \frac{1}{28} [/tex]
hence Proven
Answer:
Step-by-step 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
