Question

Those who are going to answer this question they are gonna be marked as brainlest by me but answer should be correct ​​

Answer 1

Given :-

• $\sf{{x}^{{b}^{2} + 2ca} \times {x}^{{c}^{2} + 2ab }\times {x}^{{a}^{2} + 2cb}}$

To find :-

• Simplification

Solution :-

We know that $\sf{{a}^{x} \times {a}^{y} = {a}^{X+y} }$

Using this exponential identity.

$\sf{\implies {x}^{{b}^{2} + 2ca }\times {x}^{{c}^{2} + 2ab }\times {x}^{{a}^{2} + 2cb}}$

$\sf{\implies {x}^{{b}^{2} + 2ca + {c}^{2} + 2ab + {a}^{2} + 2cb}}$

(a+b+c)² = a²+b²+c²+2ab+2cb+2ac

$\sf{\implies {x}^{{a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2cb + 2ac }}$

$\sf{\implies {x}^{{(a+b+c)}^{2}} }$

$\large{\underline{\underline{\sf{{x}^{{b}^{2} + 2ca }\times {x}^{{c}^{2} + 2ab} \times {x}^{{a}^{2} + 2cb} = {x}^{{a+b+c}^{2}} }}}}$

_____________________________

Some other identities :-

→ a³+b³ = (a+b)(a²+b²-ab)

→ a³-b³ = (a-b)(a²+b²+ab)

→ (a+b)³ = a³+b³+3ab(a+b)

→ (a-b)³ = a³-b³-3ab(a-b)

Answer 2

$\rule{200}3$

$\huge\tt{TO~FACTORISE:}$

• x^b²+2ca × x^c²+2ab × x^a²+2bc

$\rule{200}3$

$\huge\tt{</strong><strong>FACTORISING</strong><strong>:}$

Using the identity, a^x × a^y = a^x+y

↪(x^b²+2ca) × (x^c²+2ab) × (x^a²+2acb)

↪(x^b²+2ca) +(c)² +2ab + a² + 2cb

✡(a+b+c)² = a² + b² + c² + 2ab + 2cb + 2ac

↪(x^a²) + b² + c² + 2ab + 2ac + 2cb

↪(x)^(a+b+c)²

.°. x^b²+2ca × x^c²+2ab × x^a²+2bc = (x)^(a+b+c)²

$\rule{200}3$