Find the fourth proportional to :

(i) 3, 8 and 21

(ii) 1.4, 3.2 and 7

(iii) 1.5, 4.5 and 3.6

(iv) a², ab and b²

(v) (a² − ab + b²), (a³ + b³) and (a − b)

(i) Given,

3, 8 and 21

Let the fourth proportional to 3, 8 and 21 be x,

⇒ 3 : 8 = 21 : x

$\Rightarrow \dfrac{3}{8} = \dfrac{21}{x} \\[1em] \Rightarrow x = \dfrac{21 \times 8}{3} \\[1em] \Rightarrow x = \dfrac{168}{3} \\[1em] \Rightarrow x = 56.$

Hence, the fourth proportional is 56.

(ii) Given,

1.4, 3.2 and 7

Let the fourth proportional to 1.4, 3.2 and 7 be x,

⇒ 1.4 : 3.2 = 7 : x

$\Rightarrow \dfrac{1.4}{3.2} = \dfrac{7}{x} \\[1em] \Rightarrow x = \dfrac{7 \times 3.2}{1.4} \\[1em] \Rightarrow x = \dfrac{22.4}{1.4} \\[1em] \Rightarrow x = 16.$

Hence, the fourth proportional is 16.

(iii) Given,

1.5, 4.5 and 3.6

Let the fourth proportional to 1.5, 4.5 and 3.6 be x,

⇒ 1.5 : 4.5 = 3.6 : x

$\Rightarrow \dfrac{1.5}{4.5} = \dfrac{3.6}{x} \\[1em] \Rightarrow x = \dfrac{4.5 \times 3.6}{1.5} \\[1em] \Rightarrow x = 3 \times 3.6 \\[1em] \Rightarrow x = 10.8.$

Hence, the fourth proportional is 10.8.

(iv) Given,

a², ab and b²

Let the fourth proportional to a², ab and b² be x,

⇒ a² : ab = b² : x

$\Rightarrow \dfrac{a^2}{ab} = \dfrac{b^2}{x} \\[1em] \Rightarrow x = \dfrac{ab \times b^2}{a^2} \\[1em] \Rightarrow x = \dfrac{b^3}{a}.$

Hence, the fourth proportional is $\dfrac{b^3}{a}$.

(v) Given,

(a² − ab + b²), (a³ + b³) and (a − b)

Let the fourth proportional to (a² − ab + b²), (a³ + b³) and (a − b) be x,

⇒ a² : ab = b²:x

$\Rightarrow \dfrac{a^2 - ab + b^2}{a^3 + b^3} = \dfrac{a - b}{x} \\[1em] \Rightarrow x = \dfrac{(a^3 + b^3)(a - b)}{a^2 - ab + b^2} \\[1em] \Rightarrow x = \dfrac{(a + b)(a^2 - ab + b^2)(a - b)}{a^2 - ab + b^2} \\[1em] \Rightarrow x = (a + b)(a - b) \\[1em] \Rightarrow x = a^2 - b^2.$

Hence, the fourth proportional is a² - b².