Question

If x=cy + bz, y = cx + az, z = bx + ay, the value of a² +b² +c²-1 is equal to:
i)abc
ii)- abc
iii)2 abc
iv)-2 abc​

Answer 1

ANSWER:

• Option (iv) - 2abc

GIVEN:

• x = cy + bz

• y = cx + az

• z = bx + ay

TO FIND:

• The value of a² + b² + c² - 1.

EXPLANATION:

Substitute z = bx + ay in y = cx + az

y = cx + a(bx + ay)

y = cx + abx + a²y

y = (c + ab)x + a²y

Substitute z = bx + ay in x = cy + bz

x = cy + b(bx + ay)

x = cy + b²x + bay

x - b²x = cy + bay

x(1 - b²) = cy + bay

x = (cy + bay) / (1 - b²)

Substitute x = (cy + bay) / (1 - b²) in y = (c + ab)x + a²y

$\sf y = (c + ab)\left(\dfrac{cy + bay}{ 1 - b^2}\right) + a^2y$

$\sf y = y(c + ab)\left(\dfrac{c + ba}{ 1 - b^2}\right) + a^2y$

$\sf1= \dfrac{y}{y} (c + ab)\left(\dfrac{c + ba}{ 1 - b^2}\right) + \dfrac{a^2y}{y}$

$\sf 1= \dfrac{(c + ba)^{2} }{ 1 - b^2} + {a}^{2}$

$\boxed{ \bold {\large {(A + B)^2 = A^2 + 2AB + B^2}}}$

$\sf 1= \dfrac{{c}^{2} + 2abc + {a}^{2} b^{2} + {a}^{2} - {a}^{2} {b}^{2} }{ 1 - b^2}$

$\sf 1 - {b}^{2} = {c}^{2} + 2abc + {a}^{2}$

$\sf 0= {c}^{2} + 2abc + {a}^{2} + {b}^{2} - 1$

$\sf - 2abc = {a}^{2}+ {b}^{2} + {c}^{2} - 1$

Hence the value of a² + b² + c² - 1 = - 2 abc.