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prove the following [tex]\bold{algebraically}[/tex]:
[tex] \displaystyle \frac{ {29}^{3} + {29}^{2} + 30 }{ {29}^{4} - 1} = \frac{1}{28} [/tex]
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Sagot :

Nayefx

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

mathmate

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

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