Toán Lớp 8: chứng minh đẳng thức:(a+b)^3 – (a-b)^3=2b(3a^2+b)
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Toán Lớp 8: chứng minh đẳng thức:(a+b)^3 – (a-b)^3=2b(3a^2+b)
Comments ( 2 )
{\left( {a + b} \right)^3} – {\left( {a – b} \right)^3} = 2b\left( {3{a^2} + b} \right)\\
\Rightarrow {a^3} + 3{a^2}b + 3a{b^2} + {b^3} – \left( {{a^3} – 3{a^2}b + 3a{b^2} – {b^3}} \right) = 2{b^3} + 6{a^2}b\\
\Rightarrow {a^3} + 3{a^2}b + 3a{b^2} + {b^3} – {a^3} + 3{a^2}b – 3a{b^2} + {b^3} = 2{b^3} + 6{a^2}b\\
\Rightarrow 6{a^2}b + 2{b^3} = 2{b^3} + 6{a^2}b\\
\Rightarrow {\left( {a + b} \right)^3} – {\left( {a – b} \right)^3} = 2b\left( {3{a^2} + b} \right)
\end{array}\]
Ta có:
VT=(a+b)^3-(a-b)^3
VT=[(a+b)-(a-b)][(a+b)^2+(a+b)(a-b)+(a-b)^2]
VT=(a+b-a+b)(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2)
VT=[(a-a)+(b+b)][(a^2+a^2+a^2)+(b^2-b^2+b^2)+(2ab-2ab)]
VT=2b(3a^2+b^2)=VP(text{ĐPCM})
Vậy (a+b)^3-(a-b)^3=2b(3a^2+b^2)