1 способ:
По формуле суммы кубов: ![a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})](https://tex.z-dn.net/?f=a%5E%7B3%7D+%2B+b%5E%7B3%7D+%3D+%28a+%2B+b%29%28a%5E%7B2%7D+-+ab+%2B+b%5E%7B2%7D%29)
![(5d + c^{5})(25d^{2} - 5dc^{5} + c^{10}) = (5d)^{3} + (c^{5})^{3} = 125d^{3} + c^{15}](https://tex.z-dn.net/?f=%285d+%2B+c%5E%7B5%7D%29%2825d%5E%7B2%7D+-+5dc%5E%7B5%7D+%2B+c%5E%7B10%7D%29+%3D+%285d%29%5E%7B3%7D+%2B+%28c%5E%7B5%7D%29%5E%7B3%7D+%3D+125d%5E%7B3%7D+%2B+c%5E%7B15%7D)
2 способ:
Если вы не знаете формулы суммы кубов, тогда нужно раскрывать скобки:
![(5d + c^{5})(25d^{2} - 5dc^{5} + c^{10}) = 5d \ \cdotp 25 d^{2} + 5d \ \cdotp (-5dc^{5}) + 5d \ \cdotp c^{10} + c^{5} \ \cdotp 25d^{2}+ c^{5} \ \cdotp (-5dc^{5}) + c^{5} \ \cdotp c^{10} = 125d^{3} - 25d^{2}c^{5} + 5dc^{10} + 25d^{2}c^{10} - 5dc^{10} + c^{15} = 125d^{3} + c^{15}](https://tex.z-dn.net/?f=%285d+%2B+c%5E%7B5%7D%29%2825d%5E%7B2%7D+-+5dc%5E%7B5%7D+%2B+c%5E%7B10%7D%29+%3D+5d+%5C+%5Ccdotp+25+d%5E%7B2%7D+%2B+5d+%5C+%5Ccdotp+%28-5dc%5E%7B5%7D%29+%2B+5d+%5C+%5Ccdotp+c%5E%7B10%7D+%2B+c%5E%7B5%7D+%5C+%5Ccdotp+25d%5E%7B2%7D%2B+c%5E%7B5%7D+%5C+%5Ccdotp+%28-5dc%5E%7B5%7D%29+%2B+c%5E%7B5%7D+%5C+%5Ccdotp+c%5E%7B10%7D+%3D+125d%5E%7B3%7D+-+25d%5E%7B2%7Dc%5E%7B5%7D+%2B+5dc%5E%7B10%7D+%2B+25d%5E%7B2%7Dc%5E%7B10%7D+-+5dc%5E%7B10%7D+%2B+c%5E%7B15%7D+%3D+125d%5E%7B3%7D+%2B+c%5E%7B15%7D)
Ответ:![125d^{3} + c^{15}](https://tex.z-dn.net/?f=125d%5E%7B3%7D+%2B+c%5E%7B15%7D)
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4sin3x+cos²3x=0
4sin3x+1-sin²3x=0
sin²3x-4sin3x-1=0
Пусть sin3x=m, тогда
m²-4m-1=0
D=4²-4*(-1)=20
√D=2√5
m1=2+√5-пост. корень
m2=2-√5
sin3x=2-√5
3x=(-1)ⁿ*arcsin(2-√5)+Пn
x=(-1)ⁿ*arcsin(2-√5)/3+Пn/3
Мне кажется,что ответы это:
1) ZY
2)5