Factorise :
b2+c2+2bc−a2b^2 + c^2 + 2bc - a^2b2+c2+2bc−a2
1 Like
b2+c2+2bc−a2=(b2+2bc+c2)−a2=(b+c)2−a2=[(b+c)−a][(b+c)+a]=(b+c−a)(b+c+a)b^2 + c^2 + 2bc - a^2 = (b^2 + 2bc + c^2) - a^2\\[1em] = (b + c)^2 - a^2\\[1em] = [(b + c) - a][(b + c) + a]\\[1em] = (b + c - a)(b + c + a)\\[1em]b2+c2+2bc−a2=(b2+2bc+c2)−a2=(b+c)2−a2=[(b+c)−a][(b+c)+a]=(b+c−a)(b+c+a)
Hence, b2+c2+2bc−a2b^2 + c^2 + 2bc - a^2b2+c2+2bc−a2 = (b + c - a)(b + c + a).
Answered By
3 Likes
If x=1x−5x =\dfrac{1}{x-5}x=x−51, find :
(i) x−1xx-\dfrac{1}{x}x−x1
(ii) x+1xx+\dfrac{1}{x}x+x1
(iii) x2−1x2x^2-\dfrac{1}{x^2}x2−x21
(iv) x2+1x2x^2+\dfrac{1}{x^2}x2+x21
If x - y = 7 and x3−y3=133;x^3 - y^3 = 133;x3−y3=133; find :
(i) xy
(ii) x2+y2x^2 + y^2x2+y2
a2−b2−c2+2bca^2 - b^2 - c^2 + 2bca2−b2−c2+2bc
a+2b+a3+8b3a + 2b + a^3 + 8b^3a+2b+a3+8b3
