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July 2022 1 12 Report

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m11m

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А) (4x)³ +(x-1)³=(4x+x-1)((4x)² -4x*(x-1)+(x+1)²)=(5x-1)(16x²-4x²+4x+x²+2x+1)=
=(5x-1)(13x²+6x+1)

б) (3+q)³+(3+5q)³=(3+q+3+5q)((3+q)²-(3+q)(3+5q)+(3+5q)²)=
=(6+6q)(9+6q+q²-(9+3q+15q+q²)+9+30q+25q²)=
=6(1+q)(18+36q+26q²-9-18q-q²)=6(1+q)(25q²+18q+9)

в) (p-1)³+(4+4p)³=(p-1+4+4p)((p-1)²-(p-1)(4+4p)+(4+4p)²)=
=(5p+3)(p²-2p+1-(4p-4+4p²-4p)+16+32p+16p²)=
=(5p+3)(17p²+30p+17+4-4p²)=(5p+3)(13p²+30p+21)

г) (2t-1)³+(3t-1)³=(2t-1+3t-1)((2t-1)²-(2t-1)(3t-1)+(3t-1)²)=
=(5t-2)(4t²-4t+1-(6t²-3t-2t+1)+9t²-6t+1)=
=(5t-2)(13t²-10t+2-6t²+5t-1)=(5t-2)(7t²-5t+1)

д) q³p¹²a¹⁵-343r⁵¹=(qp⁴a⁵)³-(7r¹⁷)³=(qp⁴a⁵-7r¹⁷)(q²p⁸a¹⁰-7qp⁴a⁵r¹⁷+49r³⁴)

е)
$\frac{27}{64}+ \frac{125}{64}k^{102}=( \frac{3}{4} )^3+( \frac{5}{4}k^{34} )^ 3=( \frac{3}{4}+ \frac{5}{4}k^{34} )( \frac{9}{16}- \frac{15}{16}k^{34}+ \frac{25}{16}k^{68} )= \\ 
 \\ 
= \frac{1}{4}(3+5k^{34})* \frac{1}{16}*(9-15k^{34}+25k^{68})=$
$= \frac{1}{64}(3+5k^{34})(9-15k^{34}+25k^{68})$

ж)
$8f^3- \frac{1}{27}f^{21}=f^{3}(8- \frac{1}{27}f^{18} )=f^{3}(2- \frac{1}{3}f^{6} )(4+ \frac{2}{3}f^6+ \frac{1}{9}f^{12} )$

з)
$\frac{27}{8}y^6+ \frac{27}{64}h^9= \frac{27}{8}(y^6+ \frac{1}{8}h^9 )= \frac{27}{8}(y^2+ \frac{1}{2}h^3 )(y^4- \frac{1}{2}y^2h^3+ \frac{1}{4}h^6 )$

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