If a^2+1/a^2=79 where (a>0), find the value of a^3 +1/a^3

Mathematics

1 answer:

Nastasia [14]3 years ago

7 0

Note the binomial expansion,

(a + 1/a)³ = a ³ + 3a + 3/a + 1/a ³

a ³ + 1/a ³ = (a + 1/a)³ - 3 (a + 1/a)

Similarly,

(a + 1/a)² = a ² + 2 + 1/a ²

We're given a ² + 1/a ² = 79, so

(a + 1/a)² - 2 = 79

(a + 1/a)² = 81

a + 1/a = ±9

but a > 0, so we ignore the negative solution.

Then

a ³ + 1/a ³ = 9³ - 3×9 = 702

You might be interested in

A restaurant had 1/3 tray of baked pasta left over after dinner. If 6 workers share the remaining pasta equally, what fraction o

rosijanka [135]

Answer:

1/3÷6

1/3 x 1/6

Step-by-step explanation:

You want to get an answer of 1/18, which these equations give.

5 0

3 years ago

−1/3−(−1/2)

Radda [10]

Hi ;-)

$-\frac{1}{3}-(-\frac{1}{2})=-\frac{1}{3}+\frac{1}{2}=-\frac{2}{6}+\frac{3}{6}=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}\\\\3\frac{1}{3}-5=-(5-3\frac{1}{3})=-(4\frac{3}{3}-3\frac{1}{3})=-1\frac{2}{3}\\\\-1\frac{4}{5}-(-2\frac{7}{8})=-1\frac{4}{5}+2\frac{7}{8}=-1\frac{32}{40}+2\frac{35}{40}=2\frac{35}{40}-1\frac{32}{40}=1\frac{3}{40}\\\\-3\frac{3}{8}-\frac{7}{8}=-(3\frac{3}{8}+\frac{7}{8})=-3\frac{10}{8}=-3\frac{5}{4}=-4\frac{1}{4}\\\\4\frac{3}{4}-(-1\frac{1}{12})=4\frac{9}{12}+1\frac{1}{12}=5\frac{10}{12}$