Question
Simplify the following:
(2x - 5y)3 - (2x + 5y)3
(2x - 5y)3 - (2x + 5y)3
✓
Answer
Given (2x - 5y)3 - (2x + 5y)3
We shall use the identity a3 - b3 = (a - b)(a2 + b2 + ab)
Here a = (2x - 5y), b = (2x + 5y)
By applying the identity we get
$=\big(2\text{x}-5\text{y}-2\text{x}+5\text{y}\big)\\\Big[(2\text{x}-5\text{y})^2+(2\text{x}+5\text{y})^2\big((2\text{x}-5\text{y})\times(2\text{x}+5\text{y})\big)\Big]$
$=\big(2\text{x}-5\text{y}-2\text{x}-5\text{y})\Big[2\text{x}\times2\text{x}+5\text{y}\times5\text{y}-2\times2\text{x}\times5\text{y}\big)\\+\big(2\text{x}\times2\text{x}+5\text{y}\times5\text{y}+2\times2\text{x}\times5\text{y}\big)+\big(4\text{x}^2-25\text{y}^2\big)\Big]$
$=(-10\text{y})\Big[\big(4\text{x}^2+25\text{y}^2-20\text{xy}\big)\\+\big(4\text{x}^2+25\text{y}^2+20\text{xy}\big)+4\text{x}^2-25\text{y}^2\Big]$
$=(-10\text{y})\Big[4\text{x}^2+25\text{y}^2-20\text{xy}+4\text{x}^2+25\text{y}^2+20\text{xy}+4\text{x}^2-25\text{y}^2\Big]$
By rearranging the variable we get,
$=(-10\text{y})\Big[4\text{x}^2+4\text{x}^2+4\text{x}^2+25\text{y}^2\Big]$
$=-10\text{y}\times\big[12\text{x}^2+25\text{y}^2\big]$
$=-120\text{x}^2\text{y}-250\text{y}^3$
Hence the value of $\big(2\text{x}-5\text{y}\big)^3-\big(2\text{x}+5\text{y}\big)^3$ is $-120\text{x}^2\text{y}-250\text{y}^3.$
We shall use the identity a3 - b3 = (a - b)(a2 + b2 + ab)
Here a = (2x - 5y), b = (2x + 5y)
By applying the identity we get
$=\big(2\text{x}-5\text{y}-2\text{x}+5\text{y}\big)\\\Big[(2\text{x}-5\text{y})^2+(2\text{x}+5\text{y})^2\big((2\text{x}-5\text{y})\times(2\text{x}+5\text{y})\big)\Big]$
$=\big(2\text{x}-5\text{y}-2\text{x}-5\text{y})\Big[2\text{x}\times2\text{x}+5\text{y}\times5\text{y}-2\times2\text{x}\times5\text{y}\big)\\+\big(2\text{x}\times2\text{x}+5\text{y}\times5\text{y}+2\times2\text{x}\times5\text{y}\big)+\big(4\text{x}^2-25\text{y}^2\big)\Big]$
$=(-10\text{y})\Big[\big(4\text{x}^2+25\text{y}^2-20\text{xy}\big)\\+\big(4\text{x}^2+25\text{y}^2+20\text{xy}\big)+4\text{x}^2-25\text{y}^2\Big]$
$=(-10\text{y})\Big[4\text{x}^2+25\text{y}^2-20\text{xy}+4\text{x}^2+25\text{y}^2+20\text{xy}+4\text{x}^2-25\text{y}^2\Big]$
By rearranging the variable we get,
$=(-10\text{y})\Big[4\text{x}^2+4\text{x}^2+4\text{x}^2+25\text{y}^2\Big]$
$=-10\text{y}\times\big[12\text{x}^2+25\text{y}^2\big]$
$=-120\text{x}^2\text{y}-250\text{y}^3$
Hence the value of $\big(2\text{x}-5\text{y}\big)^3-\big(2\text{x}+5\text{y}\big)^3$ is $-120\text{x}^2\text{y}-250\text{y}^3.$
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$\begin{array}{c}(i)(2 a-7 b)^3=(2 a)^3+(-7 b)^3+3(2 a)(-7 b)(\_\_\_\_\_\_) \\ =\_\_\_\_\_\_+\left(-343 b^3\right)+(-42 a b)(2 a-7 b) \\ =8 a^3-\_\_\_\_\_\_\_-84 a^2 b+\_\_\_\_\_\_\end{array}$
$\begin{array}{c}(ii) 95 \times 105=(100+\_\_\_\_\_\_)\left(100-\_\_\_\_\_\_\right) \\ =(100)^2-(\_\_\_\_\_\_)^2 \\ =\_\_\_\_\_\_-\_\_\_\_\_\_\\=9975\end{array}$
→$\begin{array}{c}(i)(2 a-7 b)^3=(2 a)^3+(-7 b)^3+3(2 a)(-7 b)(\_\_\_\_\_\_) \\ =\_\_\_\_\_\_+\left(-343 b^3\right)+(-42 a b)(2 a-7 b) \\ =8 a^3-\_\_\_\_\_\_\_-84 a^2 b+\_\_\_\_\_\_\end{array}$
$\begin{array}{c}(ii) 95 \times 105=(100+\_\_\_\_\_\_)\left(100-\_\_\_\_\_\_\right) \\ =(100)^2-(\_\_\_\_\_\_)^2 \\ =\_\_\_\_\_\_-\_\_\_\_\_\_\\=9975\end{array}$