Question

2 (Q-bo2 Cath) (a+bca-b) 2 b
$(a - b)2 + (a + b)2 + (a + b)(a - b)$
​

Answer 1

Answer:

a×bb×cc×a]

(a×b).((b×c)×(c×a))

which is quarable product we know that

(a×b)×(c×d)=[abd]c−[abc]d

therefore

(a×b)([bca]c−[bcc]a

[bcc]=0 because when two roots are equal of as three

Therefore (a×b)([bca]c)

=(a×b).c[bca]

=[abc][bca]

∵[bca]=[abc]=[cab]

so, (1)[abc]

2

λ=1