Question

Prove the following alt="\bold{algebraically}" align="absmiddle" class="latex-formula">:
$\displaystyle \frac{ {29}^{3} + {29}^{2} + 30 }{ {29}^{4} - 1} = \frac{1}{28}$
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Answer 1

Answer:

see below

Step-by-step explanation:

we are given

$\displaystyle \frac{ {29}^{3} + {29}^{2} + 30 }{ {29}^{4} - 1} = \frac{1}{28}$

we want to prove it algebraically

to do so rewrite 30:

$\displaystyle \frac{ {29}^{3} + {29}^{2} + 29 + 1}{ {29}^{4} - 1} \stackrel{ ? }{= }\frac{1}{28}$

let 29 be a thus substitute:

$\displaystyle \frac{ {a}^{3} + {a}^{2} + a + 1}{ {a}^{4} - 1} \stackrel{ ? }{= }\frac{1}{28}$

factor the denominator:

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

Factor out a²:

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

factor out 1:

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

group:

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

reduce fraction:

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

$\displaystyle \frac{1}{a - 1} \stackrel {?}{ = } \frac{1}{28}$

substitute back:

$\displaystyle \frac{1}{29 - 1} \stackrel {?}{ = } \frac{1}{28}$

simplify substraction:

$\displaystyle \frac{1}{28} \stackrel { \checkmark}{ = } \frac{1}{28}$

hence Proven

Answer 2

Answer:

Solution given:

L.H.S.

$\displaystyle \frac{ {29}^{3} + {29}^{2} + 29+1 }{ {29}^{4} - 1}$

Let 29 be x.

$\displaystyle \frac{ {x}^{3} + {x}^{2} + x+1 }{ {x}^{4} - 1}$

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

$\displaystyle \frac{ (x²+1)( x+1 )}{ (x²+1)(x²-1²)}$

$\displaystyle \frac{ (x²+1)( x+1 )}{ (x²+1)(x+1)(x-1)}$

$\displaystyle \frac{1}{x-1}$

substituting value.

$\displaystyle \frac{1}{29-1}$

$\displaystyle \frac{1}{28}$

R.H.S.

proved.