Show that:
(i) $(3x + 7)^2 - 84x = (3x - 7)^2$
(ii) $(9a - 5b)^2 + 180ab = (9a + 5b)^2$
(iii) $(\frac{4m}{3} - \frac{3n}{4})^2 + 2mn = \frac{16m^2}{9} + \frac{9n^2}{16}$
(iv) $(4pq + 3q)^2 - (4pq - 3q)^2 = 48pq^2$
(v) $(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$
To do:
We have to show that:
(i) $(3x + 7)^2 - 84x = (3x - 7)^2$
(ii) $(9a - 5b)^2 + 180ab = (9a + 5b)^2$
(iii) $(\frac{4m}{3} - \frac{3n}{4})^2 + 2mn = \frac{16m^2}{9} + \frac{9n^2}{16}$
(iv) $(4pq + 3q)^2 - (4pq - 3q)^2 = 48pq^2$
(v) $(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$
Solution:
To show that LHS $=$ RHS in each case, we can use the following algebraic identities:
$(a+b)^2=a^2+2ab+b^2$.............(I)
$(a-b)^2=a^2-2ab+b^2$.............(II)
$(a+b)(a-b)=a^2-b^2$................(III)
(i) The given equation is $(3x + 7)^2 - 84x = (3x - 7)^2$
Let us consider LHS,
$(3x + 7)^2 - 84x =(3x)^2+2(3x)(7)+(7)^2-84x$ [Using (I)]
$(3x + 7)^2 - 84x =9x^2+42x+49-84x$
$(3x + 7)^2 - 84x =9x^2+42x-84x+49$
$(3x + 7)^2 - 84x =9x^2-42x+49$
$(3x + 7)^2 - 84x =(3x)^2-2(3x)(7)+(7)^2$
$(3x + 7)^2 - 84x =(3x-7)^2$ [Using (II)]
LHS $=$ RHS
Hence proved
(ii) The given equation is $(9a - 5b)^2 + 180ab = (9a + 5b)^2$
Let us consider LHS,
$(9a - 5b)^2 + 180ab =(9a)^2-2(9a)(5b)+(5b)^2+180ab$ [Using (II)]
$(9a - 5b)^2 + 180ab =(9a)^2-90ab+(5b)^2+180ab$
$(9a - 5b)^2 + 180ab =(9a)^2-90ab+180ab+(5b)^2$
$(9a - 5b)^2 + 180ab =(9a)^2+90ab+(5b)^2$
$(9a - 5b)^2 + 180ab =(9a)^2+2(9a)(5b)+(5b)^2$
$(9a - 5b)^2 + 180ab =(9a+5b)^2$ [Using (I)]
LHS $=$ RHS
Hence proved
(iii) The given equation is $(\frac{4m}{3} - \frac{3n}{4})^2 + 2mn = \frac{16m^2}{9} + \frac{9n^2}{16}$
Let us consider LHS,
$(\frac{4m}{3} - \frac{3n}{4})^2 + 2mn =(\frac{4m}{3})^2-2(\frac{4m}{3})(\frac{3n}{4})+(\frac{3n}{4})^2+2mn$ [Using (II)]
$(\frac{4m}{3} - \frac{3n}{4})^2 + 2mn =(\frac{4m}{3})^2-2mn+(\frac{3n}{4})^2+2mn$
$(\frac{4m}{3}-\frac{3n}{4})^2+2mn=\frac{(4m)^2}{3^2}-2mn+2mn+\frac{(3n)^2}{4^2}$
$(\frac{4m}{3}-\frac{3n}{4})^2+2mn=\frac{16m^2}{9}+\frac{9n^2}{16}$
LHS $=$ RHS
Hence proved
(iv) The given equation is $(4pq + 3q)^2 - (4pq - 3q)^2 = 48pq^2$
Let us consider LHS,
$(4pq + 3q)^2 - (4pq - 3q)^2 =[(4pq)^2+2(4pq)(3q)+(3q)^2]-[(4pq)^2-2(4pq)(3q)+(3q)^2]$ [Using (I) and (II)]
$(4pq + 3q)^2 - (4pq - 3q)^2 =(4pq)^2+24pq^2+(3q)^2-(4pq)^2+24pq^2-(3q)^2$
$(4pq + 3q)^2 - (4pq - 3q)^2 =(4pq)^2-(4pq)^2+24pq^2+24pq^2+(3q)^2-(3q)^2$
$(4pq + 3q)^2 - (4pq - 3q)^2 =48pq^2$
LHS $=$ RHS
Hence proved
(v) The given equation is $(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$
Let us consider LHS,
$(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) =a^2-b^2+b^2-c^2+c^2-a^2$ [Using (III)]
$(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) =a^2-a^2-b^2+b^2-c^2+c^2$
$(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) =0$
LHS $=$ RHS
Hence proved
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