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3 years ago

A^3+b^3+c^3-3abc solve this

Mathematics

1 answer:

saul85 [17]3 years ago

3 0

Answer:

Step-by-step explanation:

a3+b3+c3-3abc

Hint: a3+b3 = (a+b)3 -3ab(a+b)

(a+b)3 - 3ab(a+b) +c3 - 3abc

(a+b)3 + c3 -3ab(a+b) - 3abc

(a+b+c)3 - 3(a+b)c(a+b+c) -3ab(a + b + c)

(a+b+c)3- 3(a + b + c)[(a+b)c+ab]

(a+b+c)3 - 3(a + b + c)[(a+b)c+ab]

(a+b+c)[(a+b+c)2 -3(ab + ac + bc)]

(a+b+c)[a2 +2ab + b2 +2ac + 2bc + c2 - 3ab - 3ac- 3bc]

(a + b + c)(a2 + b2 + c2 - ab - ac- bc)

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Is the inequality always, sometimes or never true?

statuscvo [17]

In order to understand if the inequalities are always, never or sometimes true, you need to perform the calculations:

A) 9(x+2) > 9(x-3)
9x + 18 > 9x - 27
the two 9x cancel out and you get:
+18 > -27
which is always true.

B) 6x-13 < 6(x-2)
6x - 13 < 6x - 12
the two 6x cancel out and you get:
- 13 < -12
which is always true

C) -6(2x-10) + 12x ≤ 180
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All three cases are always true.

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3 years ago

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Fiesta28 [93]

Answer:

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Step-by-step explanation:

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Answer: If the level of confidence is increased without changing the sample size then The margin of error will decrease because the critical value will decrease. The decreased margin of error will cause the confidence interval to be narrower.

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3 years ago

A^3+b^3+c^3-3abc solve this​

A^3+b^3+c^3-3abc solve this