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mafiozo [28]
3 years ago
11

Expand and simplify (k-3)^7

Mathematics
2 answers:
ElenaW [278]3 years ago
4 0
(k - 3)⁷
(k - 3)(k - 3)(k - 3)(k - 3)(k - 3)(k - 3)(k - 3)
(k(k - 3) - 3(k - 3))(k(k - 3) - 3(k - 3))(k(k - 3) - 3(k - 3))(k - 3)
(k(k) - k(3) - 3(k) - 3(-3))(k(k) - k(3) - 3(k) - 3(-3))(k(k) - k(3) - 3(k) - 3(-3))(k - 3)
(k² - 3k - 3k + 9)(k² - 3k - 3k + 9)(k² - 3k - 3k + 9)(k - 3)
(k² - 6k + 9)(k² - 6k + 9)(k² - 6k + 9)(k - 3)
(k²(k² - 6k + 9) - 6k(k² - 6k + 9) + 9(k² - 6k + 9))(k²(k - 3) - 6k(k - 3) + 9(k - 3))
(k²(k²) - k²(6k) + k²(9) - 6k(k²) - 6k(-6k) - 6k(9) + 9(k²) - 9(6k) + 9(9))(k²(k) - k²(3) - 6k(k) - 6k(-3) + 9(k) - 9(3))
(k⁴ - 6k³ + 9k² - 6k³ + 36k² - 54k + 9k² - 54k + 81)(k³ - 3k² - 6k² + 18k + 9k - 27)
(k⁴ - 6k³ - 6k³ + 9k² + 36k² + 9k² - 54k - 54k - 81)(k³ - 9k² + 27k - 27)
(k⁴ - 12k³ + 54k² - 108k - 81)(k³ - 9k² + 27k - 27)
k⁴(k³ - 9k² + 27k - 27) - 12k³(k³ - 9k² + 27k² - 27) + 54k²(k³ - 9k² + 27k² - 27) - 108k(k³ - 9k² + 27k - 27) - 81(k³ - 9k² + 27k - 27)
k⁴(k³) - k⁴(9k²) + k⁴(27k) - k⁴(27) - 12k³(k³) - 12k³(-9k²) - 12k³(27k) - 12k³(-27) + 54k²(k³) - 54k²(9k²) + 54k²(27k) - 54k²(27) - 108k(k³) - 108k(9k²) - 108k(27k) - 108k(-27) - 81(k³) - 81(-9k²) - 81(27k) - 81(-27)
k⁷ - 9k⁶ + 27k⁵ - 27k⁴ - 12k⁶ + 108k⁵ - 324k⁴ + 324k³ + 54k⁵ - 486k⁴ + 1458k³ - 1458k² - 108k⁴ - 972k³ - 2916k² + 2916k - 81k³ + 729k² - 2187k + 2187
k⁷ - 9k⁶ - 12k⁶ + 27k⁵ + 108k⁵ + 54k⁵ - 27k⁴ - 324k⁴ - 486k⁴ - 108k⁴ + 324k³ + 1458k³ - 972k³ - 81k³ - 1458k² - 2196k² + 729k² + 2916k - 2187k + 2187
k⁷ - 21k⁶ + 189k⁵ - 945k⁴ + 729k³ - 2925k² + 729k + 2187
nikklg [1K]3 years ago
3 0
There are a couple of ways you can solve this. You can write it out like
(K-3)(K-3)(K-3)(K-3)(K-3)(K-3)(K-3)
And pretty much use the foil method
Or you can use the binomial theorem (google it if you dont know)
The solution should be:
(K^7) - 21(k^6) + 189(k^5) - 945(k^4) + 2835(k^3) - 5103(k^2) + 5103k - 2187
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