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Using properties of determinants prove that: |((m + n)2,l2,mn),((n + l)2,m2,ln),((l + m)2,n2,lm)| = (l - m)(m - n)(n - l)(l + m + n)(l2+ m2 + n2)|

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Best answer

We know that

So we get

On further calculation

Δ = (l2 + m2 + n2) (l – m) (m – n) (-l2 – ml + mn + n2)

It can be written as

Δ = (l2 + m2 + n2) (l – m) (m – n) (n2 – l2 + mn – ml)

Taking (n – l) as common

Δ = (l2 + m2 + n2) (l – m) (m – n) [n (n – l) + m (n – l]

We get

Δ = (l – m) (m – n) (n – l (l + m + n (l2 + m2 + n2)

Hence, it is proved.

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