Factorise p2-q2-r22-4 p2 q2A- p-q-rp-q-rp-q-r2 p-q2-rB- p-q-rp-q-rp-q-rp-q-rC- p-q-rp-q-rp-q-rp-q-rD- p-q-rp-q-rp-q-rp-q-r

(p^2+q^2 r^2)^2 4p^2q^2 = nbsp;(p^2+q^2 r^2)^2 (2pq)^2[∵ ( a^2 b^2 = (a+b)(a b)] = nbsp;(p^2+q^2 r^2 +2pq)(p^2+q^2 r^2 2pq) =(p^2+q^2+2pq r^2)(p^2+q^2 2 ...

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Byju's Answer

Standard IX

Mathematics

(a ± b)²

Factorise p2+...

Question

Factorise: $(p^{2} + q^{2} - r^{2})^{2} - 4 p^{2} q^{2}$

(p + q + r) (p + q - r) (p - q - r) (2 p - q^{2} + r)

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(p + q - r) (p + q - r) (p - q + r) (p + q + r)

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(p + q - r) (p + q - r) (p - q - r) (p - q + r)

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(p - q - r) (p + q + r) (p - q - r) (p - q - r)

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Solution

$(p^{2} + q^{2} - r^{2})^{2} - 4 p^{2} q^{2}$
$= (p^{2} + q^{2} - r^{2})^{2} - (2 p q)^{2}$
[ $∵ (a^{2} - b^{2} = (a + b) (a - b)$ ]
$= (p^{2} + q^{2} - r^{2} + 2 p q) (p^{2} + q^{2} - r^{2} - 2 p q)$
$= (p^{2} + q^{2} + 2 p q - r^{2}) (p^{2} + q^{2} - 2 p q - r^{2})$
$= ((p + q)^{2} - r^{2}) ((p - q)^{2} - r^{2})$
$= ((p + q - r) (p + q + r)) ((p - q + r) (p - q - r))$
$= (p + q - r) (p + q + r) (p - q - r) (p - q + r)$

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Question 6
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Factorise: (p2+q2−r2)2−4p2q2

(p2+q2−r2)2−4p2q2 =(p2+q2−r2)2−(2pq)2[∵(a2−b2=(a+b)(a−b)] =(p2+q2−r2+2pq)(p2+q2−r2−2pq) =(p2+q2+2pq−r2)(p2+q2−2pq−r2) =((p+q)2−r2)((p−q)2−r2) =((p+q−r)(p+q+r))((p−q+r)(p−q−r)) =(p+q−r)(p+q+r)(p−q−r)(p−q+r)

Factorise: $(p^{2} + q^{2} - r^{2})^{2} - 4 p^{2} q^{2}$