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-BARSIC- [3]
3 years ago
15

How would the expression x^3-8 be rewritten using difference of cubes ​

Mathematics
2 answers:
LUCKY_DIMON [66]3 years ago
6 0

Answer:

(x - 2)(x² + 2x + 4)

Step-by-step explanation:

A difference of cubes factors in general as

• a³ - b³ = (a - b)(a² + ab + b² )

note 8 = 2³ = 8 ⇒ b = 2 , with a = x

Hence

x³ - 8

= x³ - 2³

= (x - 2)(x² + 2x + 2²) = (x - 2)(x² + 2x + 4)

choli [55]3 years ago
3 0

Answer: The rewritten expression would be (x-2)(x^2+2x+4)

Step-by-step explanation:

Since we have given that

x^3-8

We need to use the difference of cubes:

As we know the formula of "Difference of cubes":

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

So, it can be written as

(x)^3-(2)^3=(x-2)(x^2+2x+4)

Hence, the rewritten expression would be (x-2)(x^2+2x+4)

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andrezito [222]

Answer:

a) \left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325

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Step-by-step explanation:

a) To simplify the expression \left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right) you must:

Apply Difference of Two Squares Formula: \left(a+b\right)\left(a-b\right)=a^2-b^2

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\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)

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Apply the Pythagorean Identity 1+\tan ^2\left(x\right)=\sec ^2\left(x\right)

From the Pythagorean Identity, we know that 1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)

Therefore,

324[-\tan ^2\left(x\right)+\sec ^2\left(x\right))]\\324[+1]\\325

b) According with the below graph, the lowest point of y=\cos \left(x\right), 0\leq x\leq 2\pi is when x = \pi

3 0
3 years ago
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trapecia [35]

Answer:

Rational number

Step-by-step explanation:

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7 0
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Step-by-step explanation:

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Step-by-step explanation:

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