ChatterBank6 mins ago
Maths question
Factorise a^2(b-c)+b^2(c-a)+c^2(a-b)
^2 means squared
Any ideas?
^2 means squared
Any ideas?
Answers
^ 2( b- c)+ b^ 2( c- a)+ c^ 2( a- b)= a^ 2( b- c)+ bc( b- c)- a(
b^ 2- c^ 2)
Since b^ 2- c^ 2=( b+ c)( b- c) this becomes
a^ 2( b- c)+ bc( b- c)- a( b+ c)( b- c)
=( b- c){ a^ 2+ bc- a( b+ c)}
=( b- c){ a^ 2- ab+ bc- ac}
Rearranging the stuff in the curly brackets to get a common factor of (a-c) gives
( b- c){ a( a- c)- b( a- c)}
=( b- c)( a- c)( a- b)
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08:16 Fri 11th Jun 2010
Into what form? It has already been factorised in a way, but you can reorganis eit and factorise it in another way.
First expand all the brackets
= a²b - a²c+ b²c-b²a +c²a-c²b
= a²b -b²a +c²a - a²c+ b²c -c²b
= ab(a-b) +ac(c-a) + bc (b-c)
I'm not sure this is any better though than the original expression.
Are you sure you've stated the question exactly?
First expand all the brackets
= a²b - a²c+ b²c-b²a +c²a-c²b
= a²b -b²a +c²a - a²c+ b²c -c²b
= ab(a-b) +ac(c-a) + bc (b-c)
I'm not sure this is any better though than the original expression.
Are you sure you've stated the question exactly?
If you mean factorise so it's all in brackets then you can do this (but I can't type out each line, take too long):
Multiply out first to get
a^2b - a^2c + b^2c - b^2a + c^2a - c^2b
You can then rearrange the order and get terms and take (b-c) as a factor (not doing it all for you) to give an answer of:
(b-c)(bc + a^2 - ac - ab)
Multiply out first to get
a^2b - a^2c + b^2c - b^2a + c^2a - c^2b
You can then rearrange the order and get terms and take (b-c) as a factor (not doing it all for you) to give an answer of:
(b-c)(bc + a^2 - ac - ab)