Web27 sep. 2024 · (3a + 2b -5c)³ = 0 Using the formula, (a + b + c)³ = a³ + b³ + c³+3a²b+3a²c + 3b²c +3b²a +3c²a +3c² a+6abc we get (3a + 2b -5c) ³ = 27a³ + 8b³ - 125c³ + 27a²b +27a²c + 12b²c + 12b²a + 75c²a +75c²b - 180abc = 0 ∵abc = 0 ⇒27a³ + 8b³ - 125c³ =0 hope it helped. plzz mark as brainliest Advertisement Still have questions? Find more answers WebIf a,b and c are non-coplanar vectors, then prove that the four points `2a+3b-c,a-2b+3c,3a+4b-2c Doubtnut 2.56M subscribers Subscribe 6.4K views 2 years ago If a,b and c are...
It is given that 3a + 2b= 5c. find the value of 27 a³+8b³-125c³ if …
Web6 mrt. 2024 · If a, b, c are real numbers and a2 + b2 + c2 = 2 (a - b - c) - 3, then the value of a + b + c is Q6. The value of ( 0.7) 3 − ( 0.43) 3 ( 0.7) 2 + 0.43 × 0.7 + ( 0.43) 2 is equal to: Q7. The value of ( 281 + 119) 2 − ( 281 − 119) 2 281 × 119 is equal to which of the following? Q8. If a + b 2 = 4 and ab = 7, then the value of (a - b) = Web5 apr. 2024 · If a,b,c are rational then roots of equation `abc^2x^2+3a^2 cx+b^2 cx-6a^2-ab+2b^2=0` are (A) irrational (B) rational (C) imaginary (D) irrational if `a^2ltb`. pokemon gold and silver cheat codes
Solve a^3+b^3+c^3-3abc Microsoft Math Solver
WebIf c a b c ¯ = 3 a ¯ - 2 b ¯, then prove that a b c [ a ¯ b ¯ c ¯] = 0 Advertisement Remove all ads Solution We use the results: b b 0 b ¯ × b ¯ = 0 ¯ and if in a scalar triple product, two … WebId est, it remains to prove that: a4b4c4 + abc(4(ab + ac + bc) − 9abc) ≤ 4. Let a + b + c = 3u, ab + ac + bc = 3v2 and abc = w3. Hence, the last inequality is a linear inequality of v2, which says that it's enough to prove this inequality for extremal value of v2, which happens for equality case of two variables. Web11 jul. 2024 · We have to show that roots of $$abc^2x^2 + (3a^2 c + b^2c)x-6a^2 -ab +2b^2 = 0$$ are rational. This can be possible if the discriminant is a perfect square. SO I ... pokemon gold 1997 beta version english