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If a = -3 , b= 1 C= 3, evaluate a cube b-c square -(2a-c)➗C cube -3abc?
MasiUmo10
Mar 31
If a = -3 , b= 1 C= 3, evaluate a cube b-c square -(2a-c)➗C cube -3abc?
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ChatGPT
Mar 31
First, let's substitute the given values into the expression:
a = -3
b = 1
c = 3
Now, evaluate the expression step by step:
a cube = (-3)^3 = -27
b-c square = (1-3)^2 = (-2)^2 = 4
2a-c = 2(-3) - 3 = -6 - 3 = -9
C cube = 3^3 = 27
3abc = 3(-3)(1)(3) = -27
Now substitute these values back into the expression:
-27 - 4 - (-9) / 27 - (-27)
= -27 - 4 + 9 / 27 + 27
= -22 / 54
= -11 / 27
Therefore, the final result is -11/27.
a = -3
b = 1
c = 3
Now, evaluate the expression step by step:
a cube = (-3)^3 = -27
b-c square = (1-3)^2 = (-2)^2 = 4
2a-c = 2(-3) - 3 = -6 - 3 = -9
C cube = 3^3 = 27
3abc = 3(-3)(1)(3) = -27
Now substitute these values back into the expression:
-27 - 4 - (-9) / 27 - (-27)
= -27 - 4 + 9 / 27 + 27
= -22 / 54
= -11 / 27
Therefore, the final result is -11/27.
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