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Aleks04 [339]
4 years ago
12

Explain how you can prove the difference of two cubes identity. a^3 – b^3 = (a – b)(a^2 + ab + b^2)

Mathematics
2 answers:
11111nata11111 [884]4 years ago
6 0
Is this a mutiple choice or is this a  give an example type deal

julsineya [31]4 years ago
4 0
I already calculated this

Let a^3 - b^3 = ?

Add on both th side -3a^2b + 3ab^2

a^3 -3a^2b + 3ab^2 -b^3 = ?

-3a^2b+3ab^2

The firs equation is = (a - b)^3

Then,

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

Passing -3a^2b + 3ab^2 to the left side:
But changing the sinal

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

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

Putting (a - b) as commun factor

? = (a - b).[ 3ab + (a - b)^2 ]

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

Then,

? = (a - b).[ 3ab + a^2 -2ab +b^2]

? = (a - b).( a^2 + ab + b^2)

I hope this has helped
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What is the decimal equivalent of -11/9
Aneli [31]
The decimal answer of -11/9 would be -1.22 with the .22 repeated.
4 0
3 years ago
How much must be added to each of the three numbers 1, 11, and 23 so that together they form a geometric progression?
Sliva [168]

Answer:

49

Step-by-step explanation:

Let x be unknown number which should be added to numbers 1, 11, 23 to get geometric progression. Then numbers 1 + x, 11 + x, 23 + x are first three terms of geometric progression.

Hence,

b_1=1+x\\ \\b_2=11+x\\ \\b_3=23+x

and

b_2=b_1\cdot q\Rightarrow 11+x=(1+x)q\\ \\b_3=b_2\cdot q\Rightarrow 23+x=(11+x)q

Express q:

q=\dfrac{11+x}{1+x}=\dfrac{23+x}{11+x}

Solve this equation. Cross multiply:

(11+x)^2=(1+x)(23+x)\\ \\121+22x+x^2=23+x+23x+x^2\\ \\121+22x=23+24x\\ \\22x-24x=23-121\\ \\-2x=-98\\ \\2x=98\\ \\x=49

5 0
3 years ago
Which of the following statements are true?
bezimeni [28]

Answer:

A, B, and C are all true.

Step-by-step explanation:

A. 6 is neither a perfect square nor a perfect cube.   True

B. 16 is a perfect square.   True 4^{2}

C. 27 is a perfect cube.   True 3^{3}

OD. 1,331 is both a perfect square and a perfect cube.   False

E. 9 is a perfect cube.  False

6 0
3 years ago
Approximate the square root to the nearest whole number 142
slamgirl [31]
The square root of 142 is 11.916
7 0
3 years ago
If(√14/√7-2)-(√14/√7+2)=a√7+b√2 find the values of a and b where a and b are rational numbers​
seraphim [82]

Answer:

  • a = 4/3 and b = 0

============================

<h2>Given expression:</h2>

\dfrac{\sqrt{14} }{\sqrt{7}-2} -\dfrac{\sqrt{14} }{\sqrt{7}+2}

<h2>Simplify it in steps:</h2>

<h3>Step 1</h3>

Bring both fractions into common denominator:

\dfrac{\sqrt{14} (\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} - \dfrac{\sqrt{14} (\sqrt{7}-2)}{(\sqrt{7}-2)(\sqrt{7}+2)}

<h3>Step 2</h3>

Simplify:

\dfrac{\sqrt{14} ((\sqrt{7}+2) - (\sqrt{7}-2))}{(\sqrt{7}-2)(\sqrt{7}+2)} =

\dfrac{\sqrt{14} (\sqrt{7}+2 - \sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} =

\dfrac{4\sqrt{14} }{(\sqrt{7}-2)(\sqrt{7}+2)} =

\dfrac{4\sqrt{14} }{(\sqrt{7})^2-2^2} =

\dfrac{4\sqrt{14} }{7-4} =

\dfrac{4}{3}  \sqrt{14} }

<h3>Step 3</h3>

Compare the result with given expression to get:

  • a = 4/3 and b = 0

4 0
2 years ago
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