Question

if a,b,c are real numbers such that
a-7b+8c=4 and 8a+4b-c=7, then what is the value of a²-b²+c²?
correct answer will be marked brainliest​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:a,b,c \: \in \: R$

such that

$\rm :\longmapsto\:a - 7b + 8c = 4$

$\rm :\longmapsto\:8a + 4b - c = 7$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\: \: \boxed{ \sf{ \: {a}^{2} - {b}^{2} + {c}^{2} }}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a - 7b + 8c = 4$

can be rewritten as

$\rm :\longmapsto\:a + 8c = 4 + 7b$

On squaring both sides, we get

$\rm :\longmapsto\:(a + 8c)^{2} = (4 + 7b)^{2}$

We know,

$\boxed{ \sf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

On using this identity, we get

$\rm :\longmapsto\: {a}^{2} + {64c}^{2} + 16ac = 16 + {49b}^{2} + 56b$

$\rm :\longmapsto\: {a}^{2} + {64c}^{2} - {49b}^{2} = 16 - 16ac + 56b - - (1)$

Again, Given that

$\rm :\longmapsto\:8a + 4b - c = 7$

can be rewritten as

$\rm :\longmapsto\:8a - c = 7 - 4b$

On squaring both sides, we get

$\rm :\longmapsto\:(8a - c)^{2} = (7 - 4b) ^{2}$

We know,

$\boxed{ \sf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\rm :\longmapsto\: {64a}^{2} + {c}^{2} - 16ac = 49 + {16b}^{2} - 56b$

$\rm :\longmapsto\: {64a}^{2} + {c}^{2} - {16b}^{2} = 49 + 16ac - 56b - - (2)$

Now,

Adding equation (1) and equation (2), we get

$\rm :\longmapsto\: {65a}^{2} - {65b}^{2} + {65c}^{2} = 65$

$\rm :\longmapsto\: 65({a}^{2} - {b}^{2} + {c}^{2}) \: = \: 65$

$\rm :\longmapsto\: {a}^{2} - {b}^{2} + {c}^{2}\: = \: 1$

Hence,

$\: \: \: \: \: \: \: \: \: \: \rm \rightarrow\: \: \underbrace{\boxed{ \bf{ \:{a}^{2} - {b}^{2} + {c}^{2}\: = \: 1}}} \: \: \leftarrow$

Additional Information : -

More Identities to know

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)