WebGiven, AAT = I. It represents orthogonal matrix. Determinant of orthogonal matrix is ±1. ∴ ∣A∣ = ∣∣p r q q p r r q p∣∣ = ±1. ⇒ p(p2 −qr)−q(pr −q2)+r(r2 −pq) = ±1. ⇒ p3 −pqr −pqr +q3 … WebIf p, q, r are in AP, then p 3 + r 3 - 8q 3 is equal to -6pqr. Explanation:-∵ p, q, r are in AP. ∴ 2q = p + r. ⇒ p + r – 2q = 0. ∴ p 3 + r 3 + (-2p) 3 = 3 × p × r × -2q [Using if a + 6 + c = 0 …
Ex 9.3, 5 - Add: p (p – q), q (q – r) and r (r - teachoo. Get live ...
Web5 mrt. 2024 · 3pqr(p - q)(q - r)(r - p) ... = 0. We know that when a + b + c = 0, then a³ + b³ + c³ = 3abc. ⇒ 3(pq - qr)(qr - qp)(rp - rq) ⇒ 3pqr(p - q)(q - r)(r - p) Hope it helps! Advertisement Advertisement Siddharta7 Siddharta7 if a + b + c = 0 then factors of a^3 + b^3 + c^3 = 3abc . now p(q ... WebLatest Question. Identify the pair of physical quantities which have different dimensions:Option: 1 Wave number and Rydberg's constantOption: 2 Stress and Coefficient of elasticityOption: 3 Coercivity and Magnetisation rock creek mine nome alaska
If the roots of the quadratic equation p(q - r)x^2 + q(r - p)x + r(p ...
WebIf p = −2, q = −1 and r = 3, find the value of p3 + q3 + r3 + 3pqr - Mathematics. Advertisement Remove all ads. Advertisement Remove all ads. Sum. If p = −2, q = −1 … Web2 jan. 2024 · Expanding the given expression and substituting the given values of with proves that the given equation. Correct response:. Method used to prove that the expression are equal. The given relation is;. The given equation is presented as follows;. Expanding the right hand side gives;. From the given relation, we have; p·r = q². Therefore; Web(v) pq + qr + rp, 0. Solution: (i)4p (q + r) = 4pq + 4pr. (ii)ab (a – b) = a2 b – a b2. (iii) (a + b) (7a2b2) = 7a3b2 + 7a2b3. APC Learning Mathematics - Class 8 (CBSE) - Avichal .... (i) First expression = p (p – q) = p2 – pq Second expression = q (q – r) = q2 – qr Third expression = r (r – p) = r2 – rp ∴ Required sum = (p2 – pq) + (q2 ... tes lingue online