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Mathematics

Using standard formulae, expand each of the following:

(i) (a2+b3+c4)2\Big(\dfrac{a}{2} + \dfrac{b}{3} + \dfrac{c}{4}\Big)^2

(ii) (2x3+32y−2)2\Big(\dfrac{2x}{3} + \dfrac{3}{2y} - 2\Big)^2

(iii) (2x+3x−1)2\Big(2x + \dfrac{3}{x} - 1 \Big)^2

Expansions

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Answer

We know that,

⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).

(i) Given,

⇒(a2+b3+c4)2⇒(a2)2+(b3)2+(c4)2+2×[(a2)×(b3)+(b3)×(c4)+(c4)×(a2)]⇒a24+b29+c216+2×[(ab6)+(bc12)+(ca8)]⇒a24+b29+c216+ab3+bc6+ca4\Rightarrow \Big(\dfrac{a}{2} + \dfrac{b}{3} + \dfrac{c}{4}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{a}{2}\Big)^2 +\Big(\dfrac{b}{3}\Big)^2 + \Big(\dfrac{c}{4}\Big)^2 + 2 × \Big[\big(\dfrac{a}{2}\big) × \big(\dfrac{b}{3}\big) + \big(\dfrac{b}{3}\big) × \big(\dfrac{c}{4}\big) + \big(\dfrac{c}{4}\big) × \big(\dfrac{a}{2}\big)\Big] \\[1em] \Rightarrow \dfrac{a^2}{4} + \dfrac{b^2}{9} + \dfrac{c^2}{16} + 2 × \Big[\big(\dfrac{ab}{6}\big) + \big(\dfrac{bc}{12}\big) + \big(\dfrac{ca}{8}\big)\Big] \\[1em] \Rightarrow \dfrac{a^2}{4} + \dfrac{b^2}{9} + \dfrac{c^2}{16} + \dfrac{ab}{3} + \dfrac{bc}{6} + \dfrac{ca}{4} \\[1em]

Hence, (a2+b3+c4)2=a24+b29+c216+ab3+bc6+ca4\Big(\dfrac{a}{2} + \dfrac{b}{3} + \dfrac{c}{4}\Big)^2 = \dfrac{a^2}{4} + \dfrac{b^2}{9} + \dfrac{c^2}{16} + \dfrac{ab}{3} + \dfrac{bc}{6} + \dfrac{ca}{4}.

(ii) Given,

⇒(2x3+32y−2)2⇒[2x3+32y+(−2)]2⇒(2x3)2+(32y)2+(−2)2+2×[(2x3)×(32y)+(32y)×(−2)+(−2)×(2x3)]⇒4x29+94y2+4+2×[6x6y−62y−4x3]⇒4x29+94y2+4+2×[xy−3y−4x3]⇒4x29+94y2+4+2xy−6y−8x3\Rightarrow \Big(\dfrac{2x}{3} + \dfrac{3}{2y} - 2\Big)^2 \\[1em] \Rightarrow \Big[\dfrac{2x}{3} + \dfrac{3}{2y} + (-2)\Big]^2 \\[1em] \Rightarrow \Big(\dfrac{2x}{3}\Big)^2 +\Big(\dfrac{3}{2y}\Big)^2 + (-2)^2 + 2 × \Big[\big(\dfrac{2x}{3}\big) × \big(\dfrac{3}{2y}\big) + \big(\dfrac{3}{2y}\big) × (-2) + (-2) × \big(\dfrac{2x}{3}\big)\Big] \\[1em] \Rightarrow \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + 2 × \Big[\dfrac{6x}{6y} - \dfrac{6}{2y} - \dfrac{4x}{3}\Big] \\[1em] \Rightarrow \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + 2 × \Big[\dfrac{x}{y} - \dfrac{3}{y} - \dfrac{4x}{3}\Big]\\[1em] \Rightarrow \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + \dfrac{2x}{y} - \dfrac{6}{y} - \dfrac{8x}{3} \\[1em]

Hence, (2x3+32y−2)2=4x29+94y2+4+2xy−6y−8x3\Big(\dfrac{2x}{3} + \dfrac{3}{2y} - 2\Big)^2 = \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + \dfrac{2x}{y} - \dfrac{6}{y} - \dfrac{8x}{3}.

(iii) Given,

⇒(2x+3x−1)2⇒[2x+3x+(−1)]2⇒(2x)2+(3x)2+(−1)2+2×[2x×(3x)+(3x)×(−1)+(−1)×(2x)]⇒4x2+9x2+1+2×[6−3x−2x]⇒4x2+9x2+1+12−6x−4x⇒4x2+9x2+13−6x−4x\Rightarrow \Big(2x + \dfrac{3}{x} - 1 \Big)^2 \\[1em] \Rightarrow \Big[2x + \dfrac{3}{x} + (-1) \Big]^2\\[1em] \Rightarrow (2x)^2 +\Big(\dfrac{3}{x}\Big)^2 + (-1)^2 + 2 × \Big[2x × \Big(\dfrac{3}{x}\Big) + \Big(\dfrac{3}{x}\Big) × (-1) + (-1) × (2x)\Big] \\[1em] \Rightarrow 4x^2 + \dfrac{9}{x^2} + 1 + 2 × \Big[6 - \dfrac{3}{x} - 2x\Big] \\[1em] \Rightarrow 4x^2 + \dfrac{9}{x^2} + 1 + 12 - \dfrac{6}{x} - 4x \\[1em] \Rightarrow 4x^2 + \dfrac{9}{x^2} + 13 - \dfrac{6}{x} - 4x \\[1em]

Hence, (2x+3x−1)2=4x2+9x2+13−6x−4x\Big(2x + \dfrac{3}{x} - 1 \Big)^2 = 4x^2 + \dfrac{9}{x^2} + 13 - \dfrac{6}{x} - 4x.

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