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# (4^8 - 3^8)/(4^4 + 3^4)=

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Math Expert
Joined: 02 Sep 2009
Posts: 64068
(4^8 - 3^8)/(4^4 + 3^4)=  [#permalink]

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16 Aug 2017, 01:05
00:00

Difficulty:

5% (low)

Question Stats:

88% (01:27) correct 13% (01:44) wrong based on 115 sessions

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$$\frac{{4^8 - 3^8}}{{4^4 + 3^4}}=$$

A. 7
B. 25
C. 156
D. 175
E. 216

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Intern
Joined: 22 Apr 2015
Posts: 25
WE: Business Development (Internet and New Media)
Re: (4^8 - 3^8)/(4^4 + 3^4)=  [#permalink]

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16 Aug 2017, 02:31
2
$$\frac{4^8-3^8}{4^4+3^4}$$
= $$\frac{{4^4}^{2}-{3^4}^{2}}{4^4+3^4}$$
= $$\frac{(4^4+3^4)(4^4-3^4)}{(4^4+3^4)}$$
=$$4^4-3^4$$
=256-81
=175
Director
Joined: 04 Dec 2015
Posts: 725
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)
(4^8 - 3^8)/(4^4 + 3^4)=  [#permalink]

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16 Aug 2017, 05:55
1
1
Bunuel wrote:
$$\frac{{4^8 - 3^8}}{{4^4 + 3^4}}=$$

A. 7
B. 25
C. 156
D. 175
E. 216

Can be solved using the formula : $$a^2 -b^2 = (a+b)(a-b)$$

$$\frac{{4^8 - 3^8}}{{4^4 + 3^4}}$$

$$\frac{(4^4 - 3^4)(4^4+3^4)}{(4^4 + 3^4)}$$

Cancelling $$(4^4+3^4$$); we get;

$$(4^4 - 3^4)$$

$$(4^2-3^2)(4^2 + 3^2)$$

$$(4-3)(4+3)(4^2 + 3^2)$$

$$(1)(7)(16 + 9)$$

$$(7)(25) = 175$$

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Joined: 09 Sep 2013
Posts: 14967
Re: (4^8 - 3^8)/(4^4 + 3^4)=  [#permalink]

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07 Oct 2018, 22:38
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Re: (4^8 - 3^8)/(4^4 + 3^4)=   [#permalink] 07 Oct 2018, 22:38