Question

срочно нужно ((
заранее спасибо ​

Answer 1

Ответ:

Применяем формулы куба суммы и куба разности:

             $\bf (a\pm b)^3\, =\, a^3\, \pm \, 3a^2b\, \mp \, 3ab^2\, +\, b^3$

$1)\ \ (*+4)^3=*\ +\ *\ +240+\ *\\\\(*+4)^3=(*)^3+3\cdot (*)^2\cdot 4+\underbrace{3\cdot (*)\cdot 4^2}_{240}+\underbrace{4^3}_{64}=\\\\=(*)^3+12\cdot (*)^2+\underbrace{48\cdot (*)}_{240=48\cdot 5}+64\ \ \ \ \ \ \Rightarrow \ \ \ \ \ (*)=5\\\\=5^3+12\cdot 5^2+48\cdot 5+64=125+300+240+64=729\\\\(*+4)^3=(5+4)^3=9^3=729\\\\\boldsymbol{(*+4)^3=125+300+240+64=729}$    

$2)\ \ (5+*)^3=*\ +150a^4+\ *\ +\ *\\\\(5+*)^3=5^3+\underbrace{3\cdot 5^2\cdot (*)}_{150a^2}+3\cdot 5\cdot (*)^2+(*)^3=125+\underbrace{75\cdot (*)}_{150a^2}+15\cdot (*)^2+(*)^3=\\\\\star \ \ 75\cdot (*)=150a^2=75\cdot 2a^2\ \ \ \Rightarrow \ \ \ (*)=2a^2\ \ \star \\\\=125+150a^2+15\cdot (2a^2)^2+(2a^2)^3=125+150a^2+60a^4+8a^6\\\\\bf (5+*)^3=125+150a^2+60a^4+8a^6$  

$3)\ \ (*-2z)^3=z^6\ -\ *\ +\ *\ -\ *\\\\(*-2z)^3=(*)^3-3\cdot (*)^2\cdot 2z+3\cdot (*)\cdot (2z)^2-(2z)^3=\\\\=\underbrace{(*)^3}_{z^6}-6z\cdot (*)^2+12z^2\cdot (*)-8z^3=\\\\\star \ \ (*)^3=z^6\ \ \ \Rightarrow \ \ \ \ (*)=z^2\ \ \star \\\\=(z^2)^3-6z\cdot (z^2)^2+12z^2\cdot z^2-8z^3=z^6-6z^5+12z^4-8z^3\\\\\bf (*-2z)^3=z^6-6z^5+12z^4-8z^3$  

$4)\ \ (4x^5+*)^3=*\ +\ *\ +\ *\ +\ 125x^6\\\\ (4x^5+*)^3=(4x^5)^3+3\cdot (4x^5)^2\cdot (*)+3\cdot 4x^5\cdot (*)^2+(*)^3=\\\\=64x^{15}+48x^{10}\cdot (*)+12x^5\cdot (*)^2+\underbrace{(*)^3}_{125x^6}=\\\\\star \ \ 125x^6=(*)^3\ \ \ \Rightarrow \ \ \ (*)=5x^2\ \ \star \\\\=64x^{15}+48x^{10}\cdot 5x^2+12x^5\cdot (5x^2)^2+125x^6=\\\\=64x^{15}+240x^{12}+300x^9+125x^6\\\\\bf (4x^5+*)^3=64x^{15}+240x^{12}+300x^9+125x^6$  

$5)\ \ 64b^3+\ *\ +\ *\ +125c^3=(\ *\ +\ *\ )^3\\\\64b^3+\ *\ +\ *\ +125c^3=(4b+5c)^3\ \ \ \ \Rightarrow \\\\64b^3+3\cdot (4b)^2\cdot 5c+3\cdot 4b\cdot (5c)^2+125c^3=(4b+5c)^3\\\\\bf 64b^3+240b^2c+300bc^2+125c^3=(4b+5c)^3$