Use suitable identities to find the following products: Class 9

Use suitable identities to find the following products: Class 9

Question 1.

Use suitable identities to find the following products: Class 9

i. (-3x + 4)²

ii. (2s + 7)(2s - 7)

iii. $\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)$

iv. (2n + 7)(2n - 7)

v. (s - 7t)(s² + 2st + 4t²)

vi. $\left(\frac{1}{2 r}-4 r\right)^2$

vii. (-3m + 4k - 1)²

viii. $\left(x-\frac{1}{3} y\right)^3$

ix. $\left(\frac{7}{2} k-\frac{2}{3} m\right)^3$

Solution:

i. (-3x + 4)²

Here, a = -3x, b = 4

∴ (-3x + 4)² = ( -3x)² + 2( -3x)(4) + 4²

...[∵ (a + b)² = a² + 2ab + b²]

= 9x² - 24x + 16

ii. (2s + 7)(2s - 7)

Here, a = 2s, b = 7

(2s + 7) (2s - 7) = (2s)² - 7²

...[∵ (a + b)(a -b) = a² - b²]

= 4s² - 49

iii. $\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)$

Here, a = p², b= $\frac{1}{2}$

$\begin{aligned} \therefore \quad\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right) & =\left(p^2\right)^2-\left(\frac{1}{2}\right)^2 \\ & \ldots\left[\because(a+b)(a-b)=a^2-b^2\right] \\ & =p^4-\frac{1}{4}\end{aligned}$

iv. (2n + 7)(2n - 7)

Here, a = 2n, b = 7

∴ (2n + 7)(2n - 7) = (2n)² - 7²

...[∵ (a + b)(a -b) = a² - b²]

= 4n² - 49

v. (s - 2t)(s² + 2st + 4t²)

Here, a = s, b = 2t

(s - 2t)(s² + 2st + 4t²)

= (s)³ - (2t)³ ... [∵ a³ - b³ = (a - b)(a² + ab + b²)]

= s³ - 8t³

vi. $\left(\frac{1}{2 r}-4 r\right)^2$

Here, a = $\frac{1}{2 r}$, b = 4r

$\begin{aligned} \therefore \quad\left(\frac{1}{2 r}-4 r\right)^2= & \left(\frac{1}{2 r}\right)^2-2\left(\frac{1}{2 r}\right)(4 r)+(4 r)^2 \\ & \ldots\left[\because(a-b)^2=a^2-2 a b+b^2\right] \\ = & \frac{1}{4 r^2}-4+16 r^2\end{aligned}$

vii. (-3m + 4k - l)²

Here, a = -3m, b = 4k, c = -l

(-3m + 4k - l)²

= (-3m)² + (4k)² + (-l)² + 2( -3m) (4k) + 2(4k)(-l) + 2(-l)(-3m) ... [∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

= 9m² + 16k² + l² - 24mk - 8kl + 6ml

viii. $\left(x-\frac{1}{3} y\right)^3$

Here, a = x, b= $\frac{1}{3} y$

$\begin{aligned} \therefore \quad & \left(x-\frac{1}{3} y\right)^3 \\ = & x^3-3 x^2\left(\frac{1}{3} y\right)+3 x\left(\frac{1}{3} y\right)^2-\left(\frac{1}{3} y\right)^3 \\ & \ldots\left[\because(a-b)^3=a^3-3 a^2 b+3 a b^2-b^3\right] \\ & =x^3-x^2 y+\frac{1}{3} x y^2-\frac{y^3}{27}\end{aligned}$

ix. $\left(\frac{7}{2} k-\frac{2}{3} m\right)^3$

Here, a = $\frac{7}{2} k$, b = $\frac{2}{3} m$

$\begin{aligned} \therefore \quad & \left(\frac{7}{2} k-\frac{2}{3} m\right)^3 \\ = & \left(\frac{7}{2} k\right)^3-3\left(\frac{7}{2} k\right)^2\left(\frac{2}{3} m\right) \\ & +3\left(\frac{7}{2} k\right)\left(\frac{2}{3} m\right)^2-\left(\frac{2}{3} m\right)^3 \\ & \ldots\left[\because(a-b)^3=a^3-3 a^2 b+3 a b^2-b^3\right] \\ & =\frac{343}{8} k^3-\frac{49}{2} k^2 m+\frac{14}{3} k m^2-\frac{8}{27} m^3\end{aligned}$