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Pavlova-9 [17]
3 years ago
7

Factorize 27a³-b³-1-9ab

Mathematics
2 answers:
victus00 [196]3 years ago
7 0

Answer:

Hope this helps!

Please see through the steps below

Elis [28]3 years ago
6 0

Answer:

<u>Use the following identity:</u>

  • a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c²- ab - bc - ca)

<u>Apply to the given:</u>

  • 27a³ - b³ - 1 - 9ab =
  • (3a)³ + (-b)³ + (-1)³ - 3*3a*(-b)(-1) =
  • (3a - b - 1)(9a² + b² + 1 + 3ab - b + 3a)
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Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

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Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

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\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

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