Sometimes we need to reject

Algebra Level 5

Can we say that:

if a b = c d = e f \frac { a }{ b } = \frac { c }{ d } = \frac { e }{ f } , then

a 3 b + 2 c 2 e − 3 a e 2 f b 4 + 2 d 2 f − 3 b f 3 = a c e b d f ? \frac { { a }^{ 3 }b + 2{ c }^{ 2 }e - 3a{ e }^{ 2 }f }{ { b }^{ 4 }+2{ d }^{ 2 }f-3b{ f }^{ 3 } } =\frac { ace }{ bdf } ?

Given that: a ≠ b ≠ c ≠ d ≠ e ≠ f a\neq b\neq c\neq d\neq e\neq f

False True Not enough information

Syed Baqir by Syed Baqir , 24, Cyprus

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1 solution

Syed Baqir
Aug 24, 2015

L e t a b = c d = e f = k ∴ a = b k , c = d k , e = f k S u b s i t u t i n g , ∴ a 3 b + 2 c 2 e − 3 a e 2 f b 4 + 2 d 2 f − 3 b f 3 = ( b 3 k 3 ) b + 2 ( d 2 k 2 ) ( f k ) − 3 ( b k ) ( f 2 k 2 ) f b 4 + 2 d 2 f − 3 b f 3 = k 3 ( b 4 + 2 d 2 f − 3 b f 3 ) b 4 + 2 d 2 f − 3 b f 3 = k 3 = a b ∗ c d ∗ e f = a c e b d f Let\quad \frac { a }{ b } =\frac { c }{ d } =\frac { e }{ f } =\quad k\\ \therefore \quad a\quad =\quad bk\quad ,\quad c\quad =\quad dk\quad ,\quad e\quad =\quad fk\\ Subsituting,\\ \\ \therefore \frac { { a }^{ 3 }b\quad +\quad 2{ c }^{ 2 }e\quad -\quad 3a{ e }^{ 2 }f }{ { b }^{ 4 }+2{ d }^{ 2 }f-3b{ f }^{ 3 } } =\frac { { (b }^{ 3 }{ k }^{ 3 })b\quad +\quad 2({ d }^{ 2 }{ k }^{ 2 })(fk)\quad -\quad 3(bk)({ f }^{ 2 }{ k }^{ 2 })f }{ b^{ 4 }+2{ d }^{ 2 }f-3b{ f }^{ 3 } } \\ =\frac { { k }^{ 3 }({ b }^{ 4 }+2{ d }^{ 2 }f-3b{ f }^{ 3 }) }{ { b }^{ 4 }+2{ d }^{ 2 }f-3b{ f }^{ 3 } } \quad =\quad { k }^{ 3 }\quad =\quad \frac { a }{ b } *\frac { c }{ d } *\frac { e }{ f } =\frac { ace }{ bdf }

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nope, you still have to ensure that you are not canceling out 0. E.g. if a = b = c = d = e = f = 1 a = b = c= d=e=f=1 , then the LHS is 0 0 \frac{0}{0} while the RHS is 1 1 \frac{1}{1} .

Calvin Lin Staff - 5 years, 9 months ago

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I didnot understand why we need to equate all variables to 1.

Syed Baqir - 5 years, 9 months ago

I have edited the problem thanks.

Syed Baqir - 5 years, 9 months ago

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