Using the identity #160; (a+b+c)^ 2 =a^ 2 +b^ 2 +c^ 2 +2ab+2bc+2ca .Given: #160; (p+q 2r)^2 #160;we apply the above identity as follows: (p+q ...

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Algebraic Identities

Expand: p+q...

Question

Expand: ${(p + q - 2 r)}^{2}$

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Solution

Using the identity $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ .

Given: $(p + q - 2 r)^{2}$ we apply the above identity as follows:

$(p + q + (- 2 r))^{2} = p^{2} + q^{2} + (- 2 r)^{2} + (2 \times p \times q) + (2 \times q \times (- 2 r)) + (2 \times (- 2 r) \times p) = p^{2} + q^{2} + 4 r^{2} + 2 p q - 4 q r - 4 r p$

Hence, $(p + q - 2 r)^{2} = p^{2} + q^{2} + 4 r^{2} + 2 p q - 4 q r - 4 r p$ .

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