Question

1. Prove that [a+b, b +c, c+a)= 2 [a b c].
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Answer 1

Given,

Prove that [a+b, b +c, c+a)= 2 [a b c].

To find,

We have to prove that both sides are the same.

Solution,

To prove the above equation, we must follow simple mathematic steps.

We must prove that L.H.S=R.H.S

• First, we have to solve L.H.S

⇒[a+b,b+c,c+a]

⇒(a+b)×[(b+c)×(c+a)]

By multiplying the withing brackets,

⇒(a+b)×[(b×c)+(b×a)+(c×c)+(c×a)]

• Note- c×c= |c||c| sin 0 ñ =0( assuming)

⇒(a+b)×[(b×c)+(b×a)+0+(c×a)]

⇒a×(b×c)+b×(b×c)+a×(b×a)+b×(b×a)+a×(c×a)+b×(c×a)

⇒[a,b,c]+[b²,c]+[a,b]+[b²,a]+[a²,c]+[b,c,a]

• From the following,we can say that

⇒[b²,c]=c(b×b)

⇒b×b= |b||b| sin 0 ñ =0(assuming)

⇒c×0=0.

• Now we have to apply in all the cases,

⇒[b²,c]=0,[b²,a]=0,[a²,c]=0.

• Now, the new equation is

⇒[a,b,c]+0+0+0+0+[b,c,a]

⇒We can say that bca=abc

∴[a,b,c]+[a,b,c]=2[a,b,c]

Thus R.H.S proved.

Hence, the above equation is proved (L.H.S=R.H.S).