It is true \dots or is it?

Algebra Level 2

Is a 3 = ( a ( a + 1 ) 2 ) 2 ( a ( a 1 ) 2 ) 2 a^3=\left(\dfrac{a(a+1)}{2}\right)^2-\left(\dfrac{a(a-1)}{2}\right)^2 ?

False Maybe True

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3 solutions

Chew-Seong Cheong
Sep 14, 2018

( a ( a + 1 ) 2 ) 2 ( a ( a 1 ) 2 ) 2 = a 2 4 ( ( a + 1 ) 2 ( a 1 ) 2 ) = a 2 4 ( a 2 + 2 a + 1 ( a 2 2 a + 1 ) ) = a 2 4 ( 4 a ) = a 3 True \begin{aligned} \left(\frac {a(a+1)}2\right)^2 - \left(\frac {a(a-1)}2\right)^2 & = \frac {a^2}4 ((a+1)^2 - (a-1)^2) \\ & = \frac {a^2}4 \left(a^2+2a+1 - (a^2 - 2a +1)\right) \\ & = \frac {a^2}4(4a) \\ & = a^3 \quad \boxed{\text{True}} \end{aligned}

Gia Hoàng Phạm
Sep 14, 2018

a 3 = ( a ( a + 1 ) 2 ) 2 ( a ( a 1 ) 2 ) 2 a^3=(\frac{a(a+1)}{2})^2-(\frac{a(a-1)}{2})^2

Because ( a ( a + 1 ) 2 ) 2 ( a ( a 1 ) 2 ) 2 = a 2 ( a + 1 ) 2 a 2 ( a 1 ) 2 4 = a 2 × 4 a 4 = a 3 (\frac{a(a+1)}{2})^2-(\frac{a(a-1)}{2})^2=\frac{a^2(a+1)^2-a^2(a-1)^2}{4}=\frac{a^2 \times 4a}{4}=a^3 so the answer is True

Munem Shahriar
Sep 14, 2018

{ a ( a + 1 ) 2 } 2 { a ( a 1 ) 2 } 2 = ( a 2 + a 2 ) 2 ( a 2 a 2 ) 2 = a 2 a = a 3 \begin{aligned} \left\{\dfrac {a(a+1)}2\right\}^2- \left\{\dfrac {a(a-1)}2 \right\}^2 & = \left(\dfrac{a^2+a}2\right)^2-\left(\dfrac{a^2-a}2\right)^2 \\ & = a^2 \cdot a \\ & = a^3 \\ \end{aligned}

Note: a b = ( a + b 2 ) 2 ( a b 2 ) 2 ab = \left(\dfrac{a+b}2 \right)^2 - \left(\dfrac{a-b}2 \right)^2

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