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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:
Solution given:
L.H.S.
[tex] \displaystyle \frac{ {29}^{3} + {29}^{2} + 29+1 }{ {29}^{4} - 1} [/tex]
Let 29 be x.
[tex] \displaystyle \frac{ {x}^{3} + {x}^{2} + x+1 }{ {x}^{4} - 1} [/tex]
[tex] \displaystyle \frac{ {x}^{2} (x+ 1) +1( x+1 )}{ ({x}^{2} )²- 1²} [/tex]
[tex] \displaystyle \frac{ (x²+1)( x+1 )}{ (x²+1)(x²-1²)} [/tex]
[tex] \displaystyle \frac{ (x²+1)( x+1 )}{ (x²+1)(x+1)(x-1)} [/tex]
[tex] \displaystyle \frac{1}{x-1}[/tex]
substituting value.
[tex] \displaystyle \frac{1}{29-1}[/tex]
[tex] \displaystyle \frac{1}{28}[/tex]
R.H.S.
proved.