Question

(3а-5m³)²= (0,2b²c-0,5)²= (2-е целые одна шестая а+3 в³)²= (4+х)³= (х²у-3у²)³= у4+25 а²-(у²-5а)²= (7у+5а²) (5а²-7у)= (0,4у-а³) (0,4у+а³) (а+в)³-а³-в³=

Answer 1

$(3a-5m^3)^2=(3a)^2-2*3a*5m^3+(5m^3)^2=9a^2-30am^3+25m^6$

$(0,2b^2c-0,5)^2=(0,2b^2c)^2-0,2b^2c+0,5^2=0,04b^4c^2-0,2b^2c+0,25$

$(2\frac{1}{6}a+3b^3)^2=(\frac{13}{6}a)^2+2*\frac{13}{6}a*3b^3+(3b^3)^2=\frac{169}{36}a^2+13ab^3+9b^6$

$(4+x)^3=4^3+3*4^2*x+3*4*x^2+x^3=64+48x+12x^2+x^3$
$(x^2y-3y^2)^3=(x^2y)^3-3*(x^2y)^2*3y^2+3*x^2y*(3y^2)^2-(3y^2)^3=x^6y^3-9y^2*x^4y^2+3x^2y*9y^4-27y^6=x^6y^3-9x^4y^4+27x^2y^5-27y^6=9x^2y^3(\frac{1}{9}x^4-x^2y+3y^2)-27y^6$

$y^4+25a^2-(y^2-5a)^2=y^4+25a^2-(y^4-10ay^2+25a^2)=y^4+25a^2-y^4+10ay^2-25a^2=25a^2+10ay^2-25a^2=10ay^2$

$(7y+5a^2)(5a^2-7y)=(5a^2)^2-(7y)^2=25a^4-49y^2$

$(0,4y-a^3)(0,4y+a^3)=(0,4y)^2-(a^3)^2=0,16y^2-a^6$

$(a+b)^3-a^3-b^3=a^3+3a^2b+3ab^2+b^3-a^3-b^3=3a^2b+3ab^2+b^3-b^3=3a^2b+3ab^2=3ab(a+b)$