Express `|((1+ax)^(2),(1+ay)^(2),(1+az)^(2)),((1+bx)^(2),(1+by)^(2),(1+bz)^(2)),((1+cx)^(2),(1+cy)^(2),(1+cz)^(2))|` as product of two determinants.

prove that |{:((a-x)^(2),,(a-y)^(2),,(a-z)^(2)),((b-x)^(2),,(b-y)^(2),,(b-z)^(2)),((c-x)^(2),,(c-y)^(2),,(c-z)^(2)):}| |{:((1+ax)^(2),,(1+bx)^(2),,(1+cx)^(2)),((1+ay)^(2),,(1+by)^(2),,(1+cy)^(2)),((1+az)^(2),,(1+bx)^(2),,(1+cz)^(2)):}| =2 (b-c)(c-a)(a-b)xx (y-z) (z-x)(x-y)

The value of the expression (1-ax)(1+ax)^(-1)(1+bx)^((1)/(2))(1-bx)^(-(1)/(2)) at x=a^(-1)(2(a)/(b)-1)^((1)/(2))

The value of the expression (1-ax)(1+ax)^(-1)(1+bx)^((1)/(2))(1-bx)^(-(1)/(2)) at x=a^(-1)(2(a)/(b)-1)^((1)/(2)), is: a.dependent on both a and b b.1 c.;-1

{:(ax + by = 1),(bx + ay = ((a + b)^(2))/(a^(2) + b^(2))-1):}

ax-ay=2;(a-1)x+(a+1)y=2(a^(2)+1)

If z;z_(1);z_(2)varepsilon C then (vii) |z_(1)+z_(2)|^(2)=|z_(1)|^(2)+|z_(2)|^(2)+2Re(z_(1)bar(z)_(2))( viii) |z_(1)-z_(2)|^(2)=|z_(1)|^(2)+|z_(2)|^(2)-2Re(z_(1)bar(z)_(2))( ix) |z_(1)+z_(2)|^(2)+|z_(1)-z_(2)|^(2)=2(|z_(1)|^(2)+|z_(2)|^(2))(x)|az_(1)-bz_(2)|^(2)+|bz_(1)+az_(2)|^(2)=(a^(2)+b^(2))(|z_(1)|^(2)+|z_(2)|^(2)) where a;b varepsilon R

Suppose a,b,c and x real numbers. Let Delta=|(1+a,1+ax,1+ax^(2)),(1+b,1+bx,1+bx^(2)),(1+c,1+cx,1+cx^(2))| Then Delta is independent of

ax_(1)^(2)+by_(1)^(2)+cz_(1)^(2)=ax_(2)^(2)+by_(2)^(2)+cz_(2)^(2)=ax_(3)^(2)+by_(3)^(2)+cz_(3)^(2)=d,ax_(2)^(3)+by_(2)y_(3)+cz_(2)z_(3)=ax_(3)x_(1)+by then prove that det[[x_(1),y_(1),z_(1)x_(2),y_(2),z_(2)x_(3),y_(3),z_(3)]]=(d-f){((d+2f))/(abc)}^(1/2)

If (1)/(x^(4)+x^(2)+1)=(Ax+B)/(x^(2)+x+1)+(Cx+D)/(x^(2)-x+1) then