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Find the value of the given expressions when a=0, b=-1, c=1 i) a3 – b3 ii) a2 + 2ab +b3 iii) 3ab + 3ac + c2

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Answer: (i) a^3 - b^3 = 1 when a = 0 and b = -1.
(ii) a^2 + 2ab + b^3 = -3 when a = 0 and b = -1.
(iii) 3ab + 3ac + c^2 = 1 when a = 0, b = -1, and c = 1.

Step-by-step explanation:

Sure, I'd be happy to help you with these expressions!

i) a^3 - b^3

When a = 0 and b = -1, we have:

a^3 - b^3 = 0^3 - (-1)^3 = 0 - (-1) = 1

Therefore, a^3 - b^3 = 1 when a = 0 and b = -1.

ii) a^2 + 2ab + b^3

When a = 0 and b = -1, we have:

a^2 + 2ab + b^3 = 0^2 + 2(0)(-1) + (-1)^3 = 0 - 2 + (-1) = -3

Therefore, a^2 + 2ab + b^3 = -3 when a = 0 and b = -1.

iii) 3ab + 3ac + c^2

When a = 0, b = -1, and c = 1, we have:

3ab + 3ac + c^2 = 3(0)(-1) + 3(0)(1) + 1^2 = 0 + 0 + 1 = 1

Therefore, 3ab + 3ac + c^2 = 1 when a = 0, b = -1, and c = 1.

I hope that helps! Let me know if you have any questions or if there's anything else I can do for you.

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