Prove the following - - x a a a a x a a a a x a a a a x - - x-3a x-a 3 -

Apply C _ 1 +( C _ 2 + C _ 3 + C _ 4 ) and take out x+3a common from new C_1 #160; ...

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Finding Inverse Using Elementary Transformations

Prove the fol...

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Prove the following :
$∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & a & a & a a & x & a & a a & a & x & a a & a & a & x \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = (x + 3 a) {(x - a)}^{3} .$

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∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 a b & a^{2} & b^{2} a^{2} & b^{2} & 2 a b b^{2} & 2 a b & a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = - {(a^{3} + b^{3})}^{2} .

a b c (\sum a)^{3} - (\sum b c)^{3} = a b c \sum a^{3} - \sum b^{3} c^{2} = (a^{2} - b c) (b^{2} - c a) (c^{2} - a b)

Prove the following identities :

$(i) {(a + b)}^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 b^{2} a$

$(i i) {(a - b)}^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 b^{2} a$ [4 MARKS]

Prove the following identities :

$(i) {(a + b)}^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 b^{2} a$

$(i i) {(a - b)}^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 b^{2} a$ [4 MARKS]

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