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anastassius [24]
3 years ago
15

(p²-pq + q²) ( 3p+ 3q) ​

Mathematics
1 answer:
RUDIKE [14]3 years ago
3 0

Answer:

3p^3 + qp^2 - pq^2 + 3q^3

Step-by-step explanation:

(p²-pq + q²) ( 3p+ 3q)

3p^3 + 3qp^2 - 3qp^2 - 3pq^2 + 3pq^2 + 3q^3

3p^3 + qp^2 - pq^2 + 3q^3

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3 years ago
Find the fifth roots of 243(cos 240° + i sin 240°).
Mazyrski [523]

Answer:

See below.

Step-by-step explanation:

Fifth root of 243 = 3,

Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),

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Equating equal parts and using de Moivre's theorem:

r^5 =243  and  cos  5Ф  + i sin  5Ф = cos 240 + i sin 240

r = 3 and  5Ф = 240 +360p so Ф =  48 + 72p

So Ф = 48, 120, 192, 264, 336  for   48 ≤ Ф < 360

So there are 5 distinct solutions given by:

3(cos 48 + i sin 48),

3(cos 120 + i sin 120),

3(cos 192 + i sin 192),

3(cos 264 + i sin 264),

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3 years ago
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