find 3ab+4bc+3ca-a²-b²-c²? if a=2, b=3, c=-2

asked by guest
on Nov 15, 2024 at 5:28 am



You asked:

Calculate the value of 3ab+4bc+3ca−a2−b2−c23 a b + 4 b c + 3 c a - {a}^{2} - {b}^{2} - {c}^{2} when a=2a = 2, b=3b = 3, and c=−2c = -2.

MathBot Answer:

3ab+4bc+3ca−a2−b2−c2=−353 a b + 4 b c + 3 c a - {a}^{2} - {b}^{2} - {c}^{2} = -35

−a2+3ab+3ac−b2+4bc−c2=−(−2)2−22−32+3(−2)2+3⋅3⋅2+4(−2)3=−(−2)2−32+3(−2)2+3⋅3⋅2+4(−2)3−4=−(−2)2+3(−2)2+3⋅3⋅2+4(−2)3−4−9=−(−2)2+3(−2)2+3⋅3⋅2+4(−2)3−9−4=−(−2)2+3(−2)2+3⋅3⋅2+4(−2)3−9−4=3(−2)2+3⋅3⋅2+4(−2)3−9−4−4=3(−2)2+4(−2)3−9−4+18−4=4(−2)3−12−9−4+18−4=−24−12−9−4+18−4=−35\begin{aligned}- a^{2} + 3 a b + 3 a c - b^{2} + 4 b c - c^{2}&=- \left(-2\right)^{2} - 2^{2} - 3^{2} + 3 \left(-2\right) 2 + 3 \cdot 3 \cdot 2 + 4 \left(-2\right) 3\\&=- \left(-2\right)^{2} - 3^{2} + 3 \left(-2\right) 2 + 3 \cdot 3 \cdot 2 + 4 \left(-2\right) 3 - 4\\&=- \left(-2\right)^{2} + 3 \left(-2\right) 2 + 3 \cdot 3 \cdot 2 + 4 \left(-2\right) 3 - 4 - 9\\&=- \left(-2\right)^{2} + 3 \left(-2\right) 2 + 3 \cdot 3 \cdot 2 + 4 \left(-2\right) 3 -9 - 4\\&=- \left(-2\right)^{2} + 3 \left(-2\right) 2 + 3 \cdot 3 \cdot 2 + 4 \left(-2\right) 3 -9 -4\\&=3 \left(-2\right) 2 + 3 \cdot 3 \cdot 2 + 4 \left(-2\right) 3 -9 -4 - 4\\&=3 \left(-2\right) 2 + 4 \left(-2\right) 3 -9 -4 + 18 - 4\\&=4 \left(-2\right) 3 -12 -9 -4 + 18 - 4\\&=-24 -12 -9 -4 + 18 - 4\\&=-35\end{aligned}


+ Submit Question